Multiplicativity of the Euler characteristic for fibrations For a Serre fibration
$$
F\to E \to B ,
$$
with $F,E,B$ having the homotopy type of finite complexes, it is known that the Euler characteristic is multiplicative:
$$
\chi(E) = \chi(F)\chi(B) .
$$
However, if we more generally assume that $B$ and $F$ are finitely dominated spaces, then does multiplicativity hold as well? (Recall that a finitely dominated space is a retract of a homotopy finite one.)
If true I am looking for a reference. If false, please explain.
Added later: it's true if the base is homotopy finite since we can take the fiberwise double suspension to obtain a fibration with homotopy finite fibers having the same Euler characteristic.  So we only need to consider the case when the base is a finitely dominated and the fibers are homotopy finite.
Second Addition:  I can solve the problem in general if I can solve it in the following case:  Let  $\tilde B \to B$ be a finite, regular, $n$-sheeted covering space, where $B$ is finitely dominated. Then $\chi(\tilde B) = n\chi(B)$.
(Note that $\tilde B$ is again finitely dominated, since there is a finite covering $\tilde B\times S^1 \to B\times S^1$ and by a theorem of Mather, a space $X$ is finitely dominated if and only if $X\times S^1$ is homotopy finite. But since the base $B\times S^1$ is homotopy finite, we can put a finiteness structure on the total space as well.)
Third Addition: The case alluded to in my "Second Addition" holds, by the second answer I gave below.
 A: Note Added March 1, 2022:
I now think there is a gap in deducing multiplicativity of the Euler characteristic from the Pedersen-Taylor result on the finiteness obstruction.  I think the argument I give in my other answer more-or-less fills that gap.

Ben Wieland has provided a reference which answers my question.
Pedersen, Erik Kjaer; Taylor, Lawrence R.
The Wall finiteness obstruction for a fibration.
Amer. J. Math. 100 (1978), no. 4, 887–896.
The authors identify the image in $K_0(\Bbb Z[\pi_1(B)])$ of the Wall finiteness obstruction $\sigma(E) \in K_0(\Bbb Z[\pi_1(E)])$. The Euler characteristic is the image of this class $K_0(\Bbb Z)$.
A: I have already written a lot, but Oscar Randal-Williams's comment below John Klein's second answer seems to really simplify the matter, so I decided to add an answer to that effect. I can remove any of the answers I have written if somehow they're "obfuscating the view" or anything.
This argument is just a simplification, that bypasses Pedersen-Taylor's result, and also the need for Swan's other theorem (suggested by Oscar) that $\tilde K_0(\mathbb Z[\pi])$ is finite for a finite group $\pi$ (I think the proof is harder, although I might be mistaken here). The argument is therefore due to Oscar and John in this respect, this is just my write-up of it.
Namely, as in all other answers reduce to the case of a finite covering space $p: E\to B$. Letting $F$ denote the fiber, and $Q$ the (finite) image of $\pi_1(B)\to Aut(F)$, let $f: B\to BQ$ denote the corresponding classifying map.
Define the following functor from perfect $\mathbb Z[\Omega B]$-modules to $\mathbb Q$-modules: $P\mapsto B_!( (P\otimes_\mathbb Z \mathbb Q) \otimes_\mathbb Q C_*(F;\mathbb Q))$.
The observation is that $C_*(F;\mathbb Q)$, as a $\mathbb Z[\Omega B]$-module, is of the form $f^* C_*(F;\mathbb Q)$, so this is $\mathrm{colim}_{BQ} f_!((P\otimes_\mathbb Z\mathbb Q)\otimes_\mathbb Q f^*C_*(F;\mathbb Q))\simeq \mathrm{colim}_{BQ} f_!(P\otimes_\mathbb Z\mathbb Q) \otimes C_*(F;\mathbb Q)$ (where $\mathrm{colim}_{BQ}$ is derived)
In other words, this functor factors as $Perf(\mathbb Z[\Omega B])\to Perf(\mathbb Z[Q])\to Perf(\mathbb Q[Q])\to Perf(\mathbb Q)$, but the most important part is the first part.
Indeed, $Q$ is finite, and so on $K$-theory, $K_0(\mathbb ZQ)\to K_0(\mathbb QQ)$ lands in the image of $K_0(\mathbb Q)$ (under the "induction" morphism) : this is (as far as I know) a theorem of Swan's (specifically, theorem 4.2 from his book "K-theory of finite groups and orders")
Now if $P= \mathbb Q[Q]^n$ is a free $\mathbb Q[Q]$-module, $\mathrm{colim}_{BQ}(P\otimes_\mathbb Q C_*(F;\mathbb Q)) = \mathrm{colim}_{BQ}(P\otimes_\mathbb Q C_*(F^{triv}; \mathbb Q)) = P_{hQ}\otimes_\mathbb Q C_*(F; \mathbb Q)$.
In particular, on $K$-theory, this composite has the same effect as $\mathrm{colim}_B (-)\otimes_\mathbb Q C_*(F;\mathbb Q)$, and so it sends the trivial coefficient system $\mathbb Z$ to $C_*(B;\mathbb Q)\otimes_\mathbb Q C_*(F;\mathbb Q)$, which of course is another name for $\chi(B)\chi(F) \in \mathbb Z\cong K_0(\mathbb Q)$.
But this map also sends this trivial coefficient system to $C_*(E;\mathbb Q)$, which is also another name for $\chi(E)\in \mathbb Z\cong K_0(\mathbb Q)$.
This proves the claim.
A: Here is an argument that the
Euler characteristic is multiplicative for fibrations
$$
F\to E \to B
$$
where $F$ and $B$ are finitely dominated and $B$ is connected.
Without loss in generality, we may assume that $E$ is also connected.
Step 1. The first fact is that for a homotopy pushout
$\require{AMScd}$
$$
\begin{CD}
X_{01} @>>>  X_1\\
@VVV @VVV \\
X_0 @>>> X
\end{CD}
$$
of finitely dominated spaces, the Euler characteristic is additive.  This follows from the Mayer-Vietoris sequence
using the Betti number definition of Euler characteristic. I won't include the proof here; I think the argument is easy.
Step 2. An immediate corollary is that
$$
\chi(S_B E) = 2\chi(B) - \chi(E)\, 
$$
where $S_B E\to B$ is the fiberwise suspension.  Iterating once more, we get
$$
\chi(S^2_B E) =  \chi(E)\, .
$$
But the fibers of $S^2_BE\to B$ are homotopy finite with the same Euler characteristic as
the fibers of $E\to B$, and so this enables us to reduce to the case when $E\to B$ has homotopy finite fibers.
Step 3.  The action defines a homomorphism
$$
\pi_1(B) \to \text{GL}(H_\ast(F;\Bbb Z/2))
$$
and therefore a finite covering space $\tilde B \to B$.  The base change (pullback) of the fibration $E\to B$ to $\tilde B$ yields a new fibration
$\tilde E\to \tilde B$ which is orientable in the sense of the Serre spectral sequence and the latter fibration has fiber $F$. Moreover, both $\tilde E$ and $\tilde B$ are again finitely dominated (since $B$ is and the fibers are).  It follows from Spanier's book (p. 481) that $\chi(\tilde E) = \chi(\tilde B)\chi(F)$.
Step 4. By the previous step, it is sufficient to show that the Euler characteristic is multiplicative for $n$-sheeted covering spaces
$$
E\to B
$$
with $B$ finitely dominated.  Here $F$ is a discrete set of cardinality $n$.
For this, I will appeal to Waldhausen's functor $X\mapsto A(X)$, where $A(X)$ is the finitely dominated flavor of the algebraic $K$-theory of spaces.
Step 5. Recall that $A(X)$ can be defined as the $K$-theory of the Waldhausen category of equivariant based spaces
$$
R(\ast;H)\, ,
$$
where $H$ is a topological group model for the loop space of $B$, i.e., $BH \simeq B$.  Objects of $R(\ast;H)$ are based spaces with $H$-action that are $H$-finitely dominated, in the sense that there are retracts of up to homotopy of finite objects (a finite object is one which is built out of finitely many free cells $D^k\wedge(H_+)$.)
Note that
$$
\pi_0(A(B)) = K_0(\Bbb Z[\pi_1(B)]),
$$
and the class of $S^0 \in R(\ast;H)$ is identified with the unreduced version of Wall's finiteness obstruction.
Step 6. We may identify $p:E\to B$ as induced by the homomorphism
$G\to H$ of suitable topological groups. Then one has a pushforward
and  transfer functors
$$
R(\ast;H) \overset{p^!}\to R(*;G) \overset{p_*}\to R(*;H)
$$
in which
$p^!(U) = U$ (i.e., restrict the action) and $p_*(Z) = Z\wedge_G (H_+)$.
It follows that
$$
p_*p^!(U) \cong U\wedge (F_+) ,
$$
where the target is given the diagonal $H$-action.
Claim: Let $F^t$ denote the finite set $F$ but with the trivial $H$-action (made cofibrant, i.e., $F^t = |F|\times EH$). Then $F_+,F^t_+\in R(\ast;H)$ define the same $K$-theory classes after pushforward to $R(\ast,e)$.
Sketch proof of Claim: With respect to the homomorphism $H\to e$,
the objects $F,F^t $ push forward to $E_+$ and $(B\times F)_+$
in $R(\ast;e) = R(\ast)$.
If $B$ is a finite complex, then the
Additivity Theorem and induction along the skeleta of $B$ show that
$ E_+ $ and $(B\times F)_+ $ define the same $K$-theory class.
If $B$
is finitely dominated, of dimension $r$ say, then by an argument of Wall,
there is a finite complex $K$ of dimension $r-1$ and an $(r-1)$-connected map   $K\to B$ which we can take to be a cofibration by
adjusting $B$ up to homotopy equivalence.  Furthermore, One can find such a map $K\to B$ such that
the relative homology $H_\ast(\tilde B,\tilde K;\Bbb Z)$ is trivial except in degree $r$, where $\tilde B\to B$ is the universal cover
and $\tilde K \to K$ is the pullback to $K$, and in degree $r$, we have that $P := H_r(\tilde B,\tilde K;\Bbb Z)$ is a finitely generated projective $\Bbb Z[\pi]$-module (with $\pi = \pi_1(B)$).
Note that $(-1)^rP \in K_0(\Bbb Z[\pi])$ is the Wall finiteness obstruction of $B$.
As $K$ is finite, by the above one has that $E|K$ and $K\times F$ define the same class in $\pi_0(A(\ast)) = K_0(\Bbb Z)$.
Furthermore, using the fact
that $\tilde B$ is $1$-connected, there a cofiber sequence of spaces
with $\pi$-action
$$
\tilde E_{|K} \to \tilde E \to (\tilde B/\tilde K) \wedge F_+
$$
(since the base change of $E\to B$ along $\tilde B \to B$ is a trivializable cover). Since $K$ is finite, $E_{|K}$ and $K\times F$
define the same class in $\pi_0(A(\ast))$. By additivity, it will be enough to show that
$$
H_r((\tilde B/\tilde K) \wedge_{\pi} F_+) = P\otimes_{Z[\pi]} \Bbb Z[F_+]
$$
is isomorphic to $P \otimes_{\Bbb Z[\pi]} \Bbb Z[F^t]$ as a $\Bbb  Z$-module.
To complete the proof of the claim, it is enough to establish the following:
Lemma: Suppose that $P$ is a finitely generated projective left $\Bbb Z[\pi]$ module. Then the finitely generated free $\Bbb Z$-modules
$$
P \otimes_{\Bbb Z[\pi]} \Bbb Z[F] \quad \text{ and }\quad 
P \otimes_{\Bbb Z[\pi]} \Bbb Z[F^t] 
$$
have the same rank.
Proof of Lemma  (After Oscar Randall-Williams).
In the above, to talk about the tensor product, we have regarded $P$ as a right module
using the involution $g\mapsto g^{-1}$ of $\pi$.
Let $\Pi = \text{iso}(F)$, the automorphisms of
the set $F$. Then $\Pi$  is finite. Moreover, one has a homomorphism
$\pi \to \Pi$ and we can define
$$
P' := P\otimes_{\Bbb Z[\pi]} \Bbb Z[\Pi]
$$
Then $P' \otimes_{\Bbb Z[\Pi]} \Bbb Z[F] = 
P\otimes_{\Bbb Z[\pi]} \Bbb Z[F]$ and this enables to assume at the outset that $\pi$ is finite.
So assume that $\pi$ is finite.
Consider the homomorphisms
$$
\rho,\rho^t: K_0(\Bbb Z[\pi]) \to K_0(\Bbb Z)
$$
defined by $\rho(P) = P\otimes_{\Bbb Z[\pi]} \Bbb Z[F]$, and
$\rho^t(P) = P\otimes_{\Bbb Z[\pi]} \Bbb Z[F^t]$.
It will be enough to show that these two homomorphisms coincide.
They certainly coincide for $P$ a f.g. free module of rank one and therefore for any f.g. free module.
Using Swan's Theorem that the reduced group
$\tilde K_0(\Bbb Z[\pi])$ is a torsion group (with $\pi$ finite), it follows that
$\rho = \rho^t$ in general.
$\square$.
Let $c: \pi_0(A(B)) \to \pi_0(A(\ast))$ be the map induced by the pushforward $B\to \ast$.
Hence, by the claim, the composition
$$
\begin{CD}
\pi_0(A(B)) @> p^! >> \pi_0(A(E)) @>p_\ast >> \pi_0(A(B))
@> c >> \pi_0(A(\ast)) = \Bbb Z
\end{CD}
$$
is identified with multiplication by $\chi(F)$ in the sense that
it is given by
$$
U \mapsto U_{H}\wedge F^t_+ \, .
$$
where $U_H$ is the reduced Borel construction.
On the other hand, if we take the pushforward of
$$
p_\ast p^!(S^0) = F_+
$$
in $R(\ast;e)$ (where $e$ = trivial group), we obtain $(F_+)\wedge_H *_+ = E_+$, and this will give the Euler characteristic $\chi(E) \in \pi_0(A(\ast)) = \Bbb Z$.
If we put the above facts together, we see that the composition
$$
\begin{CD}
\pi_0(A(B)) @> p_*p^!>> \pi_0(A(B)) @>c >> \pi_0(A(\ast)) = \Bbb Z 
\end{CD}
$$
has the property that it maps the class of $S^0$ to both $\chi(E)$ and $\chi(F)c([S^0])$. But $c([S^0]) = \chi(B)$, so we are done.
A: $$\newcommand{\Sph}{\mathbb S}
\newcommand{\THH}{\mathrm{THH}}
\newcommand{\Sp}{\mathrm{Sp}}$$
The answer is that Euler characteristics are multiplicative. I can't adress the question of a reference, though, because I don't know one.
Let me point out a few things: as you point out, I think the result does not quite follow from Pedersen-Taylor's paper, essentially for the reason that $\chi(p)$ is very different from $\chi(F)$ and there is a priori no reason you can "forget the $\pi_1(B)$-action" (or at least this reason should be added to the proof - but you seem to agree that this is a gap, so I won't comment further on this).
Your other answer has what I believe to be a gap, namely I think your lemma is wrong for finite groups.
Now, about the proof. I will outline a proof below that uses three ingredients : 1- a THH approach to traces; 2- the free-loop transfer and its comparison with THH transfer; 3- a result of Linnell's on the Bass trace conjecture (I learned it from a paper of Berrick-Hesselholt's, and one can make a simpler proof using TC and the Bökstedt-Hsiand-Madsen description of TC of spherical group rings). Of these, 1- is an artifact of my personal predilections, although I'm not sure how to get rid of it; 2- is not essential (I will use it in an essential way in the proof, but I also know a different, somewhat simpler argument that doesn't use it : the reason I don't outline this simpler one is that it relies on work in progress - if you are interested I would prefer to talk about it in private); and 3 is, I believe, essential, and essentially the core of the proof.
The argument goes roughly as follows: As you suggest, reduce to the case of finite Galois covers (you did not specify this, but your argument produces a Galois cover). Identify $\chi(E)$ in terms of $\THH$, and use the comparison between free loop Becker-Gottlieb transfer and THH-transfer to compare this to $\chi(F)$ : in principle, there should be some Lefschetz numbers of the monodromy action of $\pi_1(B)$ on $H_*(F)$ that appear rather than barely $\chi(F)$, but the Linnell-Hesselholt result allows you to get rid of these monodromies.
This proof may seem complicated, but I think it will be clear at the end that some complication is necessary, see the Note in Step 4. Of course it is not a precise argument, so maybe an easy argument can be found, but I think this is unlikely.
Step 0: Clearly $B$ is assumed to be connected, and we may assume that $E$ is connected too.
Step 1 : reduction to Galois covers. John already outlined this, but for convenience (and also because, even though the proof is the same, I make a stronger claim), let me recall the proof. The Euler characteristic of $F$ can be computed using $H_*(F;\mathbb Z/2)$ (or $\mathbb Z/p$ for any prime $p$), and it is easy to prove the result if the action of $\pi_1(B)$ on these is trivial, using the Serre spectral sequence. In particular, letting $\tilde B\to B$ be a finite cover corresponding to the kernel of $\pi_1(B)\to GL(H_*(F;\mathbb Z/2))$, and $\tilde E$ the pullback of $E$ to $\tilde B$, we find $\chi(\tilde E) = \chi(\tilde B)\chi(F)$, so because $\tilde E\to E, \tilde B\to B$ have the same number of sheets, it suffices to prove that for either one, $\chi(\tilde X) = n \tilde X$, $n$ being the number of sheets. Note that the kernel of $\pi_1(B)\to GL(H_*(F;\mathbb Z/2))$ is of course normal in $B$ so $\tilde B\to B$ is a Galois cover. $\tilde E\to E$ is pulled back from it, so it is Galois too; so we reduced to Galois covers.
Step 2 : $\chi(E)$ in terms of THH. Let $X$ be finite spectrum. I claim that applying $\THH$ to the morphism $\Sp^\omega\overset{X\otimes -}\to \Sp^\omega $ yields the morphism $\chi(X) : \Sph\to \Sph$ , where I use $\otimes$ for the smash product of spectra. I will abuse notation and write functors as $\Sp\to \Sp$ to avoid having to write $^\omega$ every time, but this means I will need to justify why my functors preserve compacts.
This is a fairly well-known statement, I can expend on it if needed, but I'll take it for granted for now.
Step 3 : For a finitely dominated space $X$, observe that the functor $X^* : \Sp\to \Sp^X$ taking a spectrum to the constant parametrized spectrum over $X$ with that value preserves compacts - this is precisely because $X$ is finitely dominated, so that homotopy limits over $X$ preserve filtered colimits.
It therefore induces a morphism on $\THH$, and by the Goodwillie-Jones isomorphism (I never know the actual name, someone should tell me what it is), we have $\THH( (\Sp^X)^\omega) = \THH(\mathrm{Perf}(\Sph[\Omega X])) = \THH(\Sph[\Omega X]) = \Sigma^\infty_+ LX$, where $L$ denotes the free loop space (I'm assuming $X$ is connected, to identify parametrized spectra over $X$ with $\Sph[\Omega X]$).  This morphism is therefore a morphism $\Sph\to \Sigma^\infty_+ LX$, which corresponds to an element $f_X \in \pi_0\Sigma^\infty_+ LX \cong \bigoplus_{[\gamma]\in \pi_1(X)/conj} \mathbb Z\cdot [\gamma]$, where I use brackets $[\gamma]$ to denote the conjugacy class of $\gamma$ in $\pi_1$ (or the free homotopy class of a loop).
This $f_X$ is therefore a finite sum of the form $\sum_i n_i[\gamma_i]$, where I take pairwise distinct $[\gamma_i]$'s.
Warning : It might sound "obvious" that the only $\gamma_i$ that shows up is the trivial loop. Hopefully, this is true, but it is not obvious, and as far as I know, not known so far. As I explain below, this statement is equivalent to Bass' trace conjecture.
Step 4:  Relating to Bass' trace conjecture. This $f_X$ is defined in terms of $\THH$, so one way to study it is to relate it to $K$-theory. Indeed, the trace map $K\to \THH$ is natural, so we have a commutative square $$\require{AMScd}\begin{CD}K(\Sph) @>>> K(\Sph[\Omega X]) \\
@VVV @VVV \\
\THH(\Sph) @>>> \THH(\Sph[\Omega X]) \end{CD}$$
This diagram, on $\pi_0$, becomes
$$\begin{CD}\mathbb Z = K_0(\mathbb Z) @>>> K_0(\mathbb Z[\pi_1 X]) \\
@V{id}VV @VVV \\
\mathbb Z = \THH_0(\mathbb S) @>f_X>> \bigoplus_{\pi_1(X)/conj}\mathbb Z = \mathrm{HH}_0(\mathbb Z[\pi_1(X)]\end{CD}$$
In particular, as the left vertical map is surjective, $f_X$ is in the image of the trace map $K_0(\mathbb Z[\pi_1X])\to \mathrm{HH}_0(\mathbb Z[\pi_1(X)])$. This is exactly the map that Bass' trace conjecture is about, namely:
Conjecture : Let $G$ be a group. The trace map $K_0(\mathbb Z[G])\to \mathrm{HH}_0(\mathbb Z[G]) \cong \bigoplus_{G/conj}\mathbb Z$ lands entirely in the summand corresponding to the neutral element of $G$.
Note, not necessary for the proof: The map $K_0(\mathbb S)\to K_0(\mathbb S[\Omega X])$ sends $1$ to the (unreduced) Wall finiteness obstruction of $X$. Because free modules over $\Sph[\Omega X]$ are sent to the summand of the neutral element of $\pi_1(X)$, and because any element of the reduced $K_0(\mathbb Z[\pi_1(X)])$ can be realized by some finitely dominated space, Bass' trace conjecture is equivalent to the statement that only the trivial loop shows up as one of the $\gamma_i$'s for all $X$. In particular, I will not claim that this is the case. Note, however, that if $X$ is finite, i.e. Wall's finiteness obstruction vanishes (in reduced $K$-theory), then only the trivial loop shows up.
Bass' trace conjecture is open, but some things are known about it:
Theorem ([Lin, Lemma 4.1], [BH, Theorem A]): Let $G$ be a group, and $g\in G$ an element such that $K_0(\mathbb ZG)\to \mathrm{HH}_0(\mathbb ZG)$ hits the summand corresponding to $g$. There exists an integer $m\geq 1$ such that for all $s\geq 1$, $g$ and $g^{s^m}$ are conjugate.
In particular, the $\gamma_i$'s that show up in $f_X$ have this property.
The core of the proof is in :
Corollary : Let $X$ be a finitely dominated space and write $f_X = \sum_i n_i[\gamma_i]$ as above. The $\gamma_i$'s vanish in any finite quotient of $\pi_1(X)$.
This is clear, as the image of $\gamma_i$ in the finite quotient has the same property, and has finite order.
Step 5 : A big diagram. Return to our situation: $p: E\to B$ is a finite Galois cover, with fiber $F$ a finite set.
Consider the following commutative diagram (there are diagonal arrows so I cannot use AMScd unfortunately, which is why I used a picture - hopefully it's readable enough, sorry for the inconvenience):

I use the following notation: for a map of space $f$, $f^*$ is the restriction along $f$ for parametrized spectra, and $f_!$ its (derived) left adjoint; and if $f: X\to *$ is the projection to a point, I write $X^*$ (resp. $_!$) for $f^*$ (resp. $f_!$).
From this description, it should be clear why both triangles commute. Furthermore, all functors involved preserve compact objects : indeed, their right adjoints preserve filtered colimits, either because they are of the form $f^*$, or because they are of the form $f_*$ (the right adjoint of $f^*$) for some $f$ with finitely dominated fibers.
In particular, I can apply $\THH$ to it, and get a commutative diagram.
Step 6 : The $\THH$-diagram.
It looks like (sorry, I had the same issue, I hope it's readable) :

The maps $f_E,f_B, \THH(p^*)$ are there by definition. The others follow from the following general claim:
Claim : Let $f:X\to Y$ be a map of spaces. The induced functor $f_!: \Sp^X\to \Sp^Y$ induces $\Sigma^\infty_+ Lf: \Sigma^\infty_+ LX\to \Sigma^\infty_+ LY$ on $\THH$.
This is not too hard to prove, and I think fairly well-known too. If needed, I can also add details about it (if it helps: I only need this statement on $\pi_0$ anyway !).
Important observation : the middle composite is $\THH$ applied to the middle composite, i.e. $\THH$ applied to $\Sp\overset{E^*}\to \Sp^E\overset{E_!}\to \Sp$, i.e. to $\Sp\overset{E\otimes -}\to \Sp$. Therefore, the middle composite is given by $\chi(E)$, by Step 2 above.
Step 7: Understanding $\THH(p^*)$.
To state the theorem, let me just make an observation: $p : E\to B$ is a finite covering space, therefore so is $Lp: LE\to LB$. In particular, this map has a Becker-Gottlieb transfer, which I will denote by $(Lp)^!$.
Theorem ([LM, Corollary 1.5]): Let $E\to B$ be a finite covering space. With respect to the identification $\THH((\Sp^X)^\omega)\simeq \Sigma^\infty_+ LX$, we have that $\THH(p^*)$ is identified with $(Lp)^! : \Sigma^\infty_+ LB\to \Sigma^\infty_+ LE$.
Corollary : In the above diagram, the vertical composite is $\Sigma^\infty_+ Lp\circ (Lp)^! : \Sigma^\infty_+ LB\to \Sigma^\infty_+ LB$. In particular, it is given by "multiplication by $\chi($fiber)$.
Warning : it is tempting to stop here, but as I will explain below there is a small subtlety, and this is where the Linnell-Hesselholt result comes in. A first subtlety is that in general, $LB$ is disconnected, and so the multiplication by "$\chi($fiber$)$" depends on the component under consideration.
Step 8 : Putting things together.
By the Important observation in Step 6, the middle composite of the diagram is $\chi(E)$. Because the bottom triangle commutes, one can write this composite as right-down-up instead of righ-right, and then up-down-down-up, because the top triangle commutes.
Going up gives us $f_B = \sum_i n_i[\gamma_i]$, then down-down is multiplication by $\chi(\mathrm{fib}_{\gamma_i})$, where I let $\mathrm{fib}_{\gamma_i}$ denote the fiber of $LE\to LB$ at the loop $\gamma_i$.
By Step 4, each $\gamma_i$ vanishes in $\pi_1(B)/\pi_1(E)$ (which is a finite quotient of $\pi_1(B)$, as $\pi_1(E)$ is normal in $\pi_1(B)$, and $E\to B$ is a finite cover). In particular, one can compute this fiber to be exactly $\pi_1(B)/\pi_1(E)$.
Here's the argument:
Because $E\to B$ is a Galois cover, letting $H\triangleleft G$ denote the corresponding normal inclusion, we have a homotopy pullback
$$\begin{CD}E@>>> * \\
@VVV @VVV \\
B @>>> B(G/H) \end{CD}$$
which remains a pullback after applying $L$. Now taking the fiber over $\gamma_i$, as $\gamma_i$ vanishes in $L(B(G/H))$ (more precisely: the composite $\{\gamma_i\}\to LB \to L(B(G/H))$ is homotopic to $L(*\to B(G/H))$), we can just take the outer pullback, which is $\Omega (LB(G/H), triv)$. For any group $K$, $\Omega(LBK, triv) = K$.
This proves the claim, and in particular the Euler characteristic of the fiber is $|G/H| = \chi(F)$ regardless of the $\gamma_i$, so that at the end of the day, the composite up-down-down sends $1$ to $\sum_i n_i\chi(F)[\gamma_i]$, and then going up one last time sends all loops to $1$, so up-down-down-up sends $1$ to $\sum_i n_i \chi(F)$.
So $\chi(E) = \sum_i n_i \chi(F)$. But $\sum_i n_i$ only depends on $B$ (the $n_i$'s are the ones appearing in $f_B$), so we can apply this to the cover $B\to B$, and we find $\sum_i n_i = \chi(B)$.
Therefore, $\chi(E) = \chi(B)\chi(F)$, as claimed, and we are (finally!) done.
Conclusion : The point is that in Pedersen's paper, $\chi(p)$ alone cannot be related to $\chi(F)$, it's exactly by using also $B$ that one gets to $\chi(F)$. Now this fact uses strong results about Bass' trace conjecture, and it seems like an elementary proof cannot explain this kind of "cancellation".
But maybe I am wrong. At least, I think this is a proof - of course it would be better if a more elementary one existed.
(I mentioned that I knew a simpler proof, but it still uses the Linnell-Hesselholt result)
References :
[BH] : Berrick-Hesselholt, Topological Hochschild homology and the Bass trace conjecture
[Lin] : Linnell, Decomposition of augmentation ideals and relation modules
[LM] : Lind-Malkiewich, The transfer map of free loop spaces
