Remarks:
(i) I'm interpreting the definition of $\leq_{L,\mathrm{end}}$ as quantifying over set models $W'$, not proper classes.
(ii) I'm considering the main question (comparing the two orders), particularly in the case that $V$ has large cardinals, but mainly not in the case of "particular interest", i.e. where $W$ is far from $L$; some remarks on the latter case are made in the "Edit" at the bottom.)
Claim: Assume ZF + there are ordinals $\kappa<\lambda$ such that $L_\kappa\models$ZFC and $L_\lambda\models$ZFC and $\kappa$ is a cardinal in $L_\lambda$. Let $\psi$ be the statement
"there is a countable transitive $W\models$ZFC such that defining $\leq_{L,\mathrm{end}}$ w.r.t. $W$, then $\leq_{L,\mathrm{end}}$ is different to $\leq_{\mathrm{c}}\upharpoonright W$". Then:
(i) There is a forcing extension $V[G_1]$ of $V$ such that $V[G_1]\models\psi$, and
(ii) If every real has a sharp (in particular, if there is a measurable cardinal), then $\psi$ holds in $V$.
In fact, we will get the two orders to disagree over $\mathbb{R}^W$.
Proof: The proofs of (i) and (ii) are almost the same, so I'll deal with them simulatenously.
Let $\lambda$ be least as hypothesized, and $\kappa$ the corresponding ordinal. Then $L_\lambda$ is pointwise definable by condensation etc, so $\lambda<\omega_1$,
and there is a bijection $\pi:\omega\to\lambda$ with $\pi\in L_{\lambda+2}$.
If every real has a sharp,
then we can find a sequence $\vec{y}=\left<y_\alpha\right>_{\alpha<\lambda}$ of reals such that $y_\alpha<_{\mathrm{c}}y_\beta$ for $\alpha<\beta<\lambda$, and in fact with $\vec{y}\upharpoonright \beta\in L[y_\beta]$ for each $\beta<\lambda$. In any case, we can easily force the existence of such a sequence of reals. So from now on we assume that there is such a sequence.
We will find a sequence $G=\left<x_n\right>_{n<\omega}$ which
is generic over $L_\lambda$ for the $\omega$-fold finite support product $\mathbb{C}^{<\omega}$ of Cohen forcing, such that setting $W=L_\kappa[G]$ (not $W=L_\lambda[G]$), we have $\leq_{\mathrm{c}}\upharpoonright X$ is a wellorder of ordertype $\lambda$, where $X=\{x_n\}_{n<\omega}$, and hence this order is not in $W$
(in fact not in $L_\lambda[G]$).
However, we do have $\leq_{L,\mathrm{end}}\upharpoonright X\in W$. For this, it suffices to see it is in $L_\lambda[G]$. And for this, it suffices to see it is in $L_\lambda[G,H]$ whenever $H$ is $L_\lambda[G]$-generic for $\mathrm{Coll}(\omega,\kappa)$ (by Solovay's theorem on this business; alternatively use homogeneity of the forcing and the uniformity of the next sentence). For the latter, we observe that $\leq_{L,\mathrm{end}}\upharpoonright X$ is $\Pi^1_1(\{z\})$ whenever $z$ is a real coding $L_\kappa[G]$, and since $L_\lambda[G,H]\models$ZFC and is transitive, it is therefore in $L_\lambda[G,H]$. For the $\Pi^1_1(\{z\})$-definability, it is easy assuming ZF+DC, as if there is an uncountable counterexample $W'$ to the $\forall W'$ quantifier, we can get a countable one by taking a countable hull (cf. Remark (i) at the start). Without DC we can still do a variant of this. Suppose $W'$ is a counterexample. Let $W''=L^{W'}$, and note that $W''$ is still a counterexample. Now let $W'''$ be the definable hull in $W''$ of parameters in $W\cup\{W\}$ (recalling $A,B\in W$); since $W''$ models "$V=L$", this gives an elementary substructure, and it is countable, so $W'''$ is a countable counterexample, as desired.
So we need to construct $G$.
Let $T\subseteq{^{<\omega}}2$ be a perfect tree.
Recall that $t\in T$ is a \emph{splitting node} of $T$
iff $t\frown(0)\in T$ and $t\frown(1)\in T$ (this is defined in the same manner for finite trees below). Given a real $x$,
let $b_x$ be the infinite branch
through $T$ determined by using $x(n)$ as the bit of $b_x$ following the $n$th splitting node along $b_x$. So the map ${^\omega}2\to[T]$ (the codomain is the set of branches of $T$) sending $x\mapsto b_x$ is a bijection.
Note that if $T\in L$,
then $x\equiv_{\mathrm{c}}b_x$.
Lemma: There is a perfect tree $T\subseteq {^{<\omega}}2$, with $T\in L$,
such that for every finite
sequence $\vec{b}=(b_0,\ldots,b_{m-1})$ of pairwise distinct branches $b_i$ through $T$, $\vec{b}$ is $L_\lambda$-generic
for $\mathbb{C}^m$ (the $m$-fold product of Cohen forcing).
Proof: For $m\in[1,\omega)$, let
$\mathscr{D}_m$ be the set of all
open dense $D\subseteq\mathbb{C}^m$
such that $D\in L_\lambda$.
Let $\mathscr{D}=\bigcup_{m\in[1,\omega)}\mathscr{D}_n$.
Fix an enumeration $\vec{D}=\left<D_n\right>_{n<\omega}$
of $\mathscr{D}$, such that each $D\in\mathscr{D}$ gets repeated infinitely often, and such that $\vec{D}\in L$. Let $m_n$
be the arity of $D_n$ (that is,
$D_n\subseteq\mathbb{C}^{m_n}$).
We construct a sequence $\left<T_n\right>_{n<\omega}$ of finite trees $T_n\subseteq{^{k_n}2}$
where $k_n<\omega$, such that:
(i) $T_0=\{\emptyset\}$ and $k_0=0$,
(ii) for all maximal nodes $t$ of $T_n$,
we have $\mathrm{lh}(t_n)=k_n$,
(iii) for all maximal nodes $t$ of $T_n$,
there are exactly $n$
splitting nodes $s$ of $T_n$
such that $s\subseteq t$ (so $s\subsetneq t$, since $t$ is maximal),
(iv) for each $i<n$, we have $T_i=\{t\upharpoonright k_i\bigm|t\in T_n\}$, and
each maximal node of $T_i$ is a splitting node of $T_n$ (so $k_i<k_n$), and
(v)
for each $i<n$,
letting $m=m_i$,
we have $\vec{t}\in D_i$ for each
$m$-tuple $\vec{t}=(t_0,\ldots,t_{m-1})$
of pairwise distinct maximal nodes $t_i$ of $T$.
The construction is straightforward by density: given $T_n$,
construct $T_{n+1}$ by first adding $t\frown(0)$ and $t\frown(1)$ to $T_{n+1}$ for each maximal $t$ of $T_n$, and then letting $m=m_n$,
successively extend the $m$-tuples
of distinct so-far-maximal nodes so as to get them into $D_n$. Since there are only finitely many such $m$-tuples, this is achieved after finitely many extensions, and then we just extend every node further up to some common length $k_{n+1}$ (note there the longest splitting nodes of $T_{n+1}$ are the maximal nodes of $T_n$).
Setting $T=\bigcup_{n<\omega}T_n$, we claim this works. For let $\vec{b}=(b_0,\ldots,b_{m-1})$ be an $m$-tuple of distinct branches through $T$. Let $n_0$ be large enough that $b_i\upharpoonright k_{n_0}\neq b_j\upharpoonright k_{n_0}$ whenever $0\leq i<j<m$. Then at every stage $n\in[n_0,\omega)$
such that $m_n=m$,
we ensured that $(b_0\upharpoonright k_{n+1},\ldots,b_{m-1}\upharpoonright k_{n+1})\in D_n$.
Since every open dense $D\subseteq\mathbb{C}^m$
in $L_\lambda$ gets repeated infinitely often in the enumeration,
it follows that $(b_0,\ldots,b_{m-1})$ is $L_\lambda$-generic, as desired.
This completes the proof of the lemma.
Now let $z_n=y_{\pi(n)}$ for $n<\omega$ (recall
$\left<y_\alpha\right>_{\alpha<\lambda}$ and $\pi:\omega\to\lambda$ from earlier). Let $b_n=b_{z_n}$ be the branch induced by $z_n$ (as described earlier). So ($*$) for all finite $m$-tuples of distinct integers $(n_0,\ldots,n_{m-1})$, $(b_{n_0},\ldots,b_{n_{m-1}})$ is generic over $L_\lambda$ for $\mathbb{C}^m$.
However, we don't know that the
full sequence $\left<b_n\right>_{n<\omega}$ is generic over $L_\lambda$ for the $\omega$-fold finite support product $\mathbb{C}^{<\omega}$ of $\mathbb{C}$. But by a standard trick,
using ($*$),
we can modify each $b_n$ on at most finitely many digits, producing a sequence $\left<b'_n\right>_{n<\omega}$
which is $L_\lambda$-generic for $\mathbb{C}^{<\omega}$.
(Enumerate the dense subsets of $\mathbb{C}^{<\omega}$ in $L_\lambda$
as $\left<E_n\right>_{n<\omega}$,
and progressively extend conditions
$p_n$ getting into $E_n$,
such that, letting $p_{ni}$
be the projection of $p_n$
to the $i$th component,
and letting $A_i=\mathrm{dom}(p_{ni})$,
we ensure that for $n'>n$,
we have that $p_{n'i}$ agrees
with $b_n$ outside of $A_i$.
This can be achieved using (*),
since $\mathbb{C}^{<\omega}$ factors
nicely.)
Now let $G=\left<b'_n\right>_{n<\omega}$. So $G$ is generic over $L_\lambda$ for
$\mathbb{C}^{<\omega}$, and note
that since $b'_n$ eventually agrees with $b_n$,
we have $b'_n\equiv_{\mathrm{c}} b_n\equiv_{\mathrm{c}}z_n=y_{\pi(n)}$. Therefore
letting $X=\{b'_n\}_{n<\omega}$,
the restriction $\leq_{\mathrm{c}}\upharpoonright X$ is just a wellorder of length $\lambda$,
as desired.
Edit: Consider now the case that $W$ is closed under sharps, assuming that $V$ is closed under sharps for reals. Then the preceding argument
does not work, and in fact we have the following (related to some things Hamkins wrote in his answer above):
Theorem 2: Assume ZF + for every real $x$, $x^\#$ exists.
Let $W$ be a countable transitive model of ZFC which is closed under (true) sharps.
Define $\leq_{L,\mathrm{end}}$ w.r.t. $W$.
Let $A,B\in W$.
Then $A\leq_{\mathrm{c}}B$ implies $A\leq_{L,\mathrm{end}}B$.
Proof: Suppose $A\leq_{\mathrm{c}}B$, i.e. $A\in L(B)$. Then since $B^\#\in W$,
it follows that $A\in L^W(B)=L_{\mathrm{OR}^W}(B)$. Let $W'$ be any end-extension of $W$ which models ZFC with $B\in L^{W'}$. If $\mathrm{OR}^{W'}=\mathrm{OR}^W$
then $L^{W'}=L^W$,
so $B\in L^W$, so $L^W(B)=L^W$,
so $A\in L^W=L^{W'}$. Suppose instead
that $\mathrm{OR}^W\neq\mathrm{OR}^{W'}$. Let $\gamma$ be the least $W'$-ordinal such that $B\in L^{W'}$ (note $\gamma$ might be illfounded). Let $\alpha\in\mathrm{OR}^W$ be such that $A\in L_\alpha(B)$. Then
$\alpha\in\mathrm{OR}^{W'}$,
so $\gamma+\alpha$ makes sense in $W'$ and $\gamma+\alpha\in\mathrm{OR}^{W'}$, which easily implies that $L_\alpha(B)\subseteq L^{W'}$,
so $A\in L^{W'}$, as desired.
