In a module category, the covariant Hom functors, Hom$(A,-)$, preserve direct sums when the index set is finite.
In the category of representations of a group $G$ over a field $K$, the morphisms are $K$-linear maps which are also $G$-linear. Write Hom${}_K(A,B)^G$. Since there is an isomorphism of categories, $F:K[G]-$Mod $\overset{\sim}{\rightarrow}$ Rep${}_K(G)$, I want to say that (assuming the index set is finite) we can apply $F$ to the isomorphism $$Hom_{K[G]}(A,\bigoplus_i B_i) \cong \bigoplus_i Hom_{K[G]}(A,B_i)$$ in $K[G]-$Mod to get an isomorphism in Rep${}_K(G)$: $$ Hom_{K}(A,\bigoplus_i B_i)^G \cong \bigoplus_i Hom_{K}(A,B_i)^G$$
(Since $F$ is additive, the RHS of the first isomorphism really does map to the RHS of the second.)
However, there is an issue with this reasoning: The Hom sets above may not be objects in their respective categories, only $\mathbb{Z}$-modules. For example, if $A$ is not also a right $K[G]$-module, then $Hom_{K[G]}(A,\bigoplus_i B_i)$ is not necessarily a left $K[G]$-module. So, applying $F$ to the first isomorphism doesn't always make sense.
My goal is to prove the second isomorphism above.
I've tried proving it directly, via the map $f\mapsto \sum_i p_i f$ where $p_j:\bigoplus_i B_i\rightarrow B_j$ are the natural projections. The problem here is that I can't seem to show that $p_j f\in Hom_K(A, B_j)^G$, so that this map even makes sense. Here's my attempt:
If $\rho, \theta$ are representations of $G$ corresponding to $A, \bigoplus_i B_i$, respectively, then we can define a representation $\theta_j$ of $G$ on $B_j$ by $\theta_j:=p_j \circ\theta \circ i_j$, where $i_j: B_j\rightarrow \bigoplus_i B_i$ is the natural inclusion. With this definition, let's show that $p_j f\in Hom_K(A, B_j)^G$: $$ \theta_j \circ p_j\circ f = p_j \circ \theta \circ i_j\circ p_j\circ f$$ and $$p_j \circ f\circ \rho = p_j \circ \theta\circ f.$$
So it seems that in order for these to be equal, we want $i_j\circ p_j = 1$, which is simply not true.
It seems pretty "obvious" that the second isomorphism is true. (But, obviously not to me!) Again assuming a finite index set, I'm guessing this isomorphism is also true in the other variable.
You don’t get to define $\theta_j$, since $B_j$ is already a represention of $G$. Currently you’re doing the following:
Taking a sum $\bigoplus_i B_i$ of vector spaces $B_i$.
Choosing the structure of a $G$-representation on $\bigoplus_i B_i$, named $\theta$.
For every $i$, restrict the structure $\theta$ to a the structure of a $G$-representation on $B_i$, named $\theta_i$.
Try to show that $\bigoplus_i B_i$ is the direct sum of the $B_i$ as representations.
The problem with this is that we can not necessarily restrict $\theta$ to a $G$-representation on $B_i$. This works if and only if $B_i$ is a subrepresentation.
Take for example $I = \{1,2\}$, $B_1 = B_2 = K$. Then $G = \mathbb{Z}/2$ acts on $B_1 \oplus B_2 = K^2$ by switching the coordinates. But this action can not be restricted to $B_1$ and $B_2$.
This problem is solved by the fact that $\theta$ is defined through the $\theta_j$ (and not vice versa, as attempted above): If $\theta_j$ is the $G$-representation on $B_j$, then the $G$-representation $\theta$ on $\bigoplus_i B_i$ is (by definition of the direct sum of representations $\bigoplus_i B_i$) given by $\theta = \bigoplus_i \theta_i$. The inclusions $i_j$ and projections $p_j$ are then homomorphisms of representations, because $$ p_j \circ \theta = p_j \circ \left( \bigoplus_i \theta_i \right) = \theta_j \circ p_j $$ and similarly $\theta \circ i_j = i_j \circ \theta_j$. We then get that $$ \theta_j \circ p_j \circ f = p_j \circ \theta \circ f = p_j \circ f \circ \rho. $$
I also want to point out that for an arbitrary index set $J$ we have that \begin{align*} \operatorname{Hom}\left( A, \prod_{j \in J} B_j \right)^G \cong \prod_{j \in J} \operatorname{Hom}(A, B_j)^G, \quad f \mapsto (p_j \circ f)_{j \in I} \end{align*} and $$ \operatorname{Hom}\left( \bigoplus_{j \in J} A_j, B \right)^G \cong \prod_{j \in J} \operatorname{Hom}(A_j, B)^G, \quad f \mapsto (f \circ i_j)_{j \in I} $$ as $K$-vector spaces. It may actually be more enlightening to check these two isomorphisms, and to check things on elements instead of trying to do everying in a purely abstract way.