Let $G$ be a group acting on (complex) vector spaces $V, W$ and let $G$ act trivially on the vector space $U$.
Let $Hom^G(V, W)$ denote the linear transformations $T:V \to W$ that respect the group action, that is, $g \cdot T(v) = T(g \cdot v)$ for all $v \in V$.
$W \otimes U$ has a natural $G$-action given by $g \cdot (w \otimes u) = gw \otimes gu = gw \otimes u$.
How do I see $Hom^G(V, W \otimes U) \cong Hom^G(V, W) \otimes U$?
There is a natural map going backwards given by sending $(T: V \to W) \otimes u$ to the map $v \mapsto T(v) \otimes u$. This second map respects the $G$-action because $T$ respects the $G$-action, and $U$ has trivial $G$-action.
I don't believe this is true in full generality; one needs $U$ to be finite-dimensional.
In the case where $G$ is the trivial group, this amounts to $$\text{Hom}(V,W\otimes U)\cong\text{Hom}(V,W)\otimes U.$$ Specialising further to $W=\Bbb C$ (one-dimensional) we get $$\text{Hom}(V,U)\cong\text{Hom}(V,\Bbb C)\otimes U.\tag{*}$$ The natural map from the right of (*) to the left has its image the homomorphisms from $V$ to $U$ with finite-dimensional image, so is not surjective.
But if $U$ is finite-dimensional one can write $U$ as finite direct sum of trivial one-dimensional modules. As both functors $U\mapsto \text{Hom}^G(V,W\otimes U)$ and $U\mapsto \text{Hom}^G(V,W)\otimes U$ preserve finite direct sums, we reduce to the case $U=\Bbb C$ with trivial $G$-action, in which case the result is immediate.