In the case that $V$ and $W$ are finite-dimensional, here is a "non-intrinsic" way to see this isomorphism. If $V$ is finite-dimensional, then every $\operatorname{End}(V)$-module is semisimple, and moreover every simple $\operatorname{End}(V)$-module is isomorphic to $V$:
Choose a basis $\{e_i\}$ for $V$ and define $P_i(e_j)=\delta_{ij}e_i$, so $P_1,\ldots,P_n\in\operatorname{End}(V)$ with $P_iP_j=\delta_{ij}P_i$ and $\sum_iP_i=1$. Each of the left modules $\operatorname{End}(V)\cdot P_i$ is then simple and isomorphic to $V$, via the map $A\cdot P_i\mapsto Ae_i$. This is easier to see if you think of this choice of basis as giving an isomorphism between $\operatorname{End}(V)$ and the ring of $n\times n$ matrices over your field, in which case $\operatorname{End}(V)\cdot P_i$ is the submodule of matrices whose only nonzero column is the $i$th one.
Let $M$ be a simple $\operatorname{End}(V)$-module and let $x\in M\setminus\{0\}$. Since $\sum_iP_i=1$, there must be some $i$ for which $P_ix\neq 0$, hence $M=\operatorname{End}(V)\cdot P_ix$. The map of modules $\operatorname{End}(V)\cdot P_i\to M$ sending $aP_i$ to $aP_ix$ is then an isomorphism, by Schur's lemma.
Combining this with semisimplicity, if $W$ is finite-dimensional then it is isomorphic as an $\operatorname{End}(V)$-module to some direct sum $V^{\oplus n}$.
If we unravel the chain of isomorphisms $\operatorname{Hom}_{\operatorname{End}(V)}(V,V^{\oplus n})\otimes V\cong \operatorname{Hom}_{\operatorname{End}(V)}(V,V)^{\oplus n}\otimes V\cong (\operatorname{Hom}_{\operatorname{End}(V)}(V,V)\otimes V)^{\oplus n}\cong (\mathbb{C}\otimes V)^{\oplus n}\cong V^{\oplus n}$, we see that it is the same as the evaluation map you define.