Since i could not find the reference book, below the following earlier version of it is used:

CQGRC.pdf - Compact Quantum Groups and Their Representation Categories - Incomplete preliminary version as of January 17, 2014, by Sergey Neshveyev and Lars Tuset.

In [CQGRC], it is Proposition: 1.7.2, CQGRC, page 21 claiming that $f_z$ has the property (i) of being a is a homomorphism $\Bbb C[G]\to\Bbb C$. and the proof of (i) claims the relation $$ \tag{$\dagger$} \hat\Delta(\rho^z)=\rho^z\otimes \rho^z\ . $$ So let us comment first on relation $(\dagger)$ above, which is the question. For the convenience of the reader, we unpack the notations one by one, and describe the framework:

Framework:

$G=(A,\Delta)$ is the given $C^*$-algebraic CQG. The underlying algebra is in an alternative notation $A=C(G)$. (I will use preferably $A$ below, since it is simpler to type.)

We need the space of matrix coefficients of $A$, denoted alternatively by $A_0=\Bbb C[G]$.

The functionals $f_z$ are linear functionals $f_z:A_0\to \Bbb C$, and we write $A_0'$ for this space of functionals. So for each complex $z$ we have $f_z\in A_0'$. To put the hands on some $f_z$ we need to know the structure of $A_0$.

What is $A_0$? The first proposition in CQGRC, §1.6, is:

For a compact quantum group $G=(A,\Delta)$ denote by $A_0=\Bbb C[G]\subset C(G)=A$ the linear span of matrix coefficients of all finite dimensional (co)representations of $G$, it is a dense subspace of $A$ by CQGRC, Corollary 1.5.6. So let $(U_\alpha)$ be such a maximal family of irreducible, unitary, mutually inequivalent representations of $G=(A,\delta)$. The index $\alpha$ runs in some index set $\Lambda$, that we will often omit. Here, $$ U_\alpha = \sum m^\alpha_{ij}\otimes u^\alpha_{ij}\in B(H_\alpha)\otimes A\ . $$ The matrices $m^\alpha_{ij}$ are the elementary matrices, the canonical basis of the matrix space $B(H_\alpha)$. For an irreducible, unitary representation $U\in B(H) \otimes A_0$ its contragredient cousin is $U^c$. CQGRC, Proposition 1.3.13 shows that $U$ and $U^{cc}$ are equivalent.

The space of intertwiners between $U$, and $U^{cc}$, in notation $\operatorname{Mor}(U,U^{cc})\operatorname{Mor}(U,U)$ is thus one-dimensional (Schur), and we have even an explicitly constructed intertwiner $j(Q_r)\in \operatorname{Mor}(U, U^{cc})$. Here $Q_r$ is a positive, invertible operator in $B(\bar H)$, and $j$ is the antimorphism on $B(H)$. We rescale $Q_r$ by a positive scalar $\lambda_U$, so that the obtained invertible, positive operator $\rho_U:=\lambda_U\; j(Q_r)$ is normed by the condition: $$ \operatorname{Trace}(\rho_U) = \operatorname{Trace}(\rho_U^{-1}) \ . $$ From now on, we can forget about $j(Q_r)$, and $\lambda_U$, and need only the information that the space of intertwiners is one-dimensional, generated by an invertible, positive operator $\rho_U$.

Since $\rho_U>0$, and for each $z$ the map $a\to a^z:=e^z\log a$ is analytic on $(0,\infty)$, we have an analytic functional calculus defining the positive operator $$(\rho_U)^z=:\rho_U^z\in B(H)\ .$$

Now, $f_z$ is introduced in CQGRC, Definition 1.7.1, implicitly. We do not have a formula to evaluate $f_z$ on an elements from $A_0$, instead, we use the matrix coefficient spaces of individual irreducible representations.
Note that for different (inequivalent) unitary (co)representations $U_\alpha$, $U_\beta$ the corresponding spaces of matrix coefficients

• $\mathscr C(\alpha):=\operatorname{Span}(U_\alpha):=\operatorname{Span}(u_{\alpha,ij})$ and

• $\mathscr C(\beta):=\operatorname{Span}(U_\beta):=\operatorname{Span}(u_{\beta,kl})$

generated by their entries are linearly independent (orthogonal). Moreover, this independence holds also for the coefficients of $U_\alpha$, so that extending $f_z$ is uniquely defined on $\mathscr C(\alpha)$ by the relation: $$ %\Big(\operatorname{id}_{B(H_{\alpha})}\otimes f_z\Big)\;\Big(U_\alpha\Big) \ =\ \rho_{\alpha}^z\in B(H_{\alpha})_{>0}\ . (\operatorname{id}\otimes f_z)(U_\alpha)\ =\ \rho_{\alpha}^z\in B(H_{\alpha})\ . $$ We denoted by $\rho_\alpha$ the positive operator $\rho_{H_\alpha}$ constructed above. (Unpacking, if $m_{\alpha,ij}$ is the canonical basis of $B(H_\alpha)$, and $U_\alpha =\sum m^\alpha_{ij}\otimes u^\alpha_{ij}$ to obtain $f_z(u^\alpha_{ij})$ we compute $\rho_\alpha$, then its $z$-power by analytic functional calculus, then we write this result in the $m^\alpha$--basis and take the coefficient in $m^\alpha_{ij}$.

So far we have the $\rho_U$ objects. Then the $\rho_U^z$ objects. Using them we get $f_z\in A_0'$. We forget about them, and need in fact only $f_1$. So we need in fact only $\rho_U$ to know $f_1$ on the vector space spanned by matrix coefficients of $U$. And now, the text also uses the letter $\rho$ for $f_1$. (This may be nice a posteriori. But at this point, let us be more careful.)

So i will slightly change notation and use instead: $$ \varrho:=f_1\in A_0'=\mathscr U(G)\overset{\Phi=\Phi_G}\longrightarrow \prod_\alpha B(H_\alpha)\ . $$ The text identifies $\varrho$ with its image $\Phi(\varrho)$ in the product of matrix spaces. Let us compute it explicitly, using $\pi_{U_\alpha}$ as in CQGRC, page 20: $$ \begin{aligned} \Phi(\varrho) &=(\ \Phi(\varrho)_\alpha\ )_\alpha \ ,\\ \Phi(\varrho)_\alpha &:= \pi_{U_\alpha}(\varrho)\\ &:=(\operatorname{id}\otimes\varrho)(U_\alpha)\\ &:=(\operatorname{id}\otimes f_1)(U_\alpha)\\ &:=\rho_{U_\alpha}^1=\rho_{U_\alpha}=\rho_\alpha\ . \end{aligned} $$ In other words, $\Phi(\varrho)$ is an element in $\prod B(H_\alpha)$ which has as $\alpha$-component the positive operator $\rho_\alpha\in B(H_\alpha)$. I will write $\rho$ for this family, $\rho=(\rho_\alpha)=\Phi(\varrho)$.

(At any rate, we can now imagine why the notation $\rho$ was kept for $\varrho$.)

What is $\varrho^z\in \mathscr U(G)=A_0'$? It is by definition the functional which is mapped by $\Phi$ into $\rho^z:=(\rho_\alpha^z)_\alpha$.

So identities involving $\varrho^z$ should be rather moved from the $\mathscr U$-spaces to the matrix spaces.

Let us recall $\hat\Delta$ is as a map $\mathscr{U}(G)\to \mathscr{U}(G\times G)$, which is determined by its evaluation at some functional $\omega \in A_0' =\mathscr{U}(G)$. The image lies in $\mathscr{U}(G\times G)$. (It can be bigger than $A_0'\otimes A_0'$.) The functional $\hat \Delta(\omega)\in \mathscr{U}(G\times G)$ is determined by evaluation on elements of the shape $a\otimes b\in A_0\otimes A_0$. So $\hat\Delta$ is determined by the double evaluation $$ \hat\Delta(\omega)\ a\otimes b :=\omega(ab)\ . $$ Our $\omega$ of interest is $\rho^z$. So it is enough to show the above relation for two elements of the shape $a=g(U_\alpha)$, $b=h(U_\beta)$ with $g,h\in A_0'$: $$ \tag{$1$} \hat\Delta(\rho^z)\ g(U_\alpha) :=\omega(ab)\ . $$ Recall that we also have $$ \mathscr U(G)\overset{\Phi_G}\longrightarrow \prod_{\alpha} B(H_\alpha) $$ given by mapping $\omega\in \mathscr U(G)=A_0'$ to $(\ (\operatorname{id}_{H_\alpha}\otimes\omega)(U_\alpha)\ )_\alpha$.

The same $\Phi$-mapping can be written also for $G\times G$, the corresponding irreducible representations are parametrized by tuples $(\alpha,\beta)$ and are of the shape $U_\alpha\odot U_\beta\in B(H_{(\alpha,\beta)})\otimes A\otimes A$, where $B(H_{(\alpha,\beta)}):=B(H_\alpha)\otimes B(H_\beta)$ it the algebraic tensor product of the two matrix spaces. Then consider the diagram, which is for a general $\omega$ not commutative: $$ \require{AMScd} \begin{CD} \omega @>\hat\Delta_G>> \hat\Delta(\omega) \mathscr U(G) @>\hat\Delta_G>> \mathscr U(G\times G)\\ @V \Phi_G V V (??) @VV \Phi_{G\times G} V\\ \prod_\alpha B(H_\alpha) @>>\underline\Delta > \prod_{(\alpha,\beta)} B(H_\alpha) \otimes B(H_\beta) (w_\alpha)_{\alpha\in\Lambda} @>>\underline\Delta > (w_\alpha\otimes w_\beta)_{(\alpha,\beta)\in\Lambda\times\Lambda} \end{CD} $$ For the special value $\omega=\varrho$, which is a "group-like" element, $\hat\Delta(\varrho)=\varrho\otimes \varrho$ we have the commutativity $$ \require{AMScd} \begin{CD} \varrho @>\hat\Delta >> \hat\Delta(\varrho)\\ @V \Phi_G V V (??) @VV \Phi_{G\times G} V\\ \rho=(\rho_\alpha) @>>\underline\Delta > \rho\otimes \rho =(\rho_\alpha\otimes\rho_\beta)_{(\alpha,\beta)} \end{CD} $$ by using CQGRC, Theorem 1.4.8, $$ \rho_{U_\alpha\times U_\beta}=\rho_{U_\alpha}\otimes \rho_{U_\beta}\ . $$ Well, this is as stated an equality in the category of representations for $G$, however the part from $A$ involved in what we need is evaluated to a factor.

So we compute the component $(\alpha,\beta)$ of $\hat\Delta$ evaluated in $\varrho$. We write explicitly at some point:

• $U_\alpha =\sum m^\alpha_{ij}\otimes u^\alpha_{ij}\in B(H_\alpha)\otimes A_0$, representation of $G$,
• $U_\beta =\sum m^\beta_{kl}\otimes u^\beta_{kl}\in B(H_\beta)\otimes A_0$, representation of $G$, so that
• $U_{(\alpha,\beta)}=\sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes u^\alpha_{ij}\otimes u^\beta_{kl}\in B(H_\alpha)\otimes B(H_\beta)\otimes A_0\otimes A_0$

is the corresponding tensor product representation of $G\times G$. (Not for $G$, as in the notation above.)

$$ \begin{aligned} (\ \Phi_{G\times G}\ \hat\Delta(\varrho)\ )_{(\alpha,\beta)} &:= \big(\ \operatorname{id}_{B(H_{(\alpha,\beta)})}\otimes\hat\Delta(\varrho)\ \Big)(U_{(\alpha,\beta)}) \\ &= \big(\ \operatorname{id}_{B(H_{\alpha})}\otimes \operatorname{id}_{B(H_{\beta})}\otimes \hat\Delta(\varrho)\ \Big) \sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes u^\alpha_{ij}\otimes u^\beta_{kl} \\ &= \sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes \hat\Delta(\varrho)(u^\alpha_{ij}\otimes u^\beta_{kl}) \\ &= \sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes \hat\Delta(\varrho)(u^\alpha_{ij}\otimes u^\beta_{kl}) \\ &= \sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes \underbrace{(\varrho\otimes\varrho)(u^\alpha_{ij}\otimes u^\beta_{kl})}_{\in\Bbb C} \\ &= \sum m^\alpha_{ij}\otimes m^\beta_{kl}\cdot \varrho(u^\alpha_{ij}\cdot \varrho(u^\beta_{kl}) \\ &= \sum m^\alpha_{ij}\cdot \varrho(u^\alpha_{ij}\otimes \sum m^\beta_{kl}\cdot \varrho(u^\beta_{kl}) \\ &= (\operatorname{id}\otimes\varrho)\left(\sum m^\alpha_{ij}\otimes u^\alpha_{ij}\right) \otimes (\operatorname{id}\otimes\varrho)\left(\sum m^\beta_{kl}\otimes u^\beta_{kl}\right) \\ &= (\operatorname{id}\otimes\varrho)(U_\alpha) \otimes (\operatorname{id}\otimes\varrho)(U_\beta) \\ &=\rho_\alpha\otimes\rho_beta \\ &=\underline\Delta(\rho)_{(\alpha,\beta)} \\ &=\underline\Delta(\Phi(\varrho))_{(\alpha,\beta)} \ . \end{aligned} $$ As a final work, a way of giving a sense to the boxed identity from the question, $$ \hat\Delta(\varrho^z) =\varrho^z\otimes\varrho^z $$ which is related to the upper horizontal arrow in the above diagrams, is by moving it downwards via the $\Phi$ arrows to the lower horizontal arrow, which is a map $\underline\Delta$ clearly compatible with the functional calculus, $$ \underline\Delta(\rho^z)_{(\alpha,\beta)} = \rho_\alpha^z\otimes\rho_\beta^z = (\rho_\alpha\otimes\rho_\beta)^z = (\ \underline\Delta(\rho)\ )^z \ . $$ The definition of $\varrho^z$ is by taking $\rho^z$ from the L-most.H.S. and pushing it via $\Phi^{-1}$ into $\mathscr U(G)$. It may be then useful in the vertical $G\times G$-arrow to write $\varrho^z\otimes\varrho^z$ as a product of $\varrho^z\otimes\epsilon$ and $\epsilon\otimes\varrho^z$, then go down via $\Phi$ to get by definition of $\varrho$ the commuting operators $\rho^z\otimes 1$ and $1\otimes \rho^z$, and here we have $$ (\rho\otimes\rho)^z = (\ (\rho\otimes 1)\;(1\otimes\rho)\ )^z = (\rho\otimes 1)^z\;(1\otimes\rho)^z\ . $$

The notations in CQGRC were my biggest problems. I hope the above answer - a general nonsense categorial translation - hits the wound point.