Finding induced isomorphism from $\DeclareMathOperator{\Hom}{Hom} \Hom_F(U\oplus U_2, V)\to \Hom_f(U_, V)\oplus \Hom_f(U_2, V)$ I am seeing this kind of question a lot in my class, but each time fail to understand how to think about the question. Why is there an isomorphism between sets of homomorphisms and what is it explicitly? 
 A: Let $R$ be a ring and $U_1, U_2, V$ be $R$-modules.  First, recall that we have canonical injections into the coproduct:
\begin{align*}
\iota_1: U_1 &\hookrightarrow U_1 \oplus U_2\\
u_1 &\mapsto (u_1, 0)\\
\iota_2: U_2 &\hookrightarrow U_1 \oplus U_2\\
u_2 &\mapsto (0, u_2) \, .
\end{align*}
Given $R$-linear maps $\varphi_1: U_1 \to V$ and $\varphi_2: U_2 \to V$, the universal property of the coproduct states that there is a unique $R$-linear map $\varphi = \varphi_1 \oplus \varphi_2: U_1 \oplus U_2 \to V$ such that the following diagram commutes.
$\hskip1.9in$
Commutativity of the diagram implies that $(\varphi_1 \oplus \varphi_2)(u_1, u_2) = \varphi_1(u_1) + \varphi_2(u_2)$.
Conversely, given an $R$-linear map $\varphi: U_1 \oplus U_2 \to V$, pre-composition by the injections yields maps $U_1 \to V$ and $U_2 \to V$.  Explicitly, let $\varphi_1 = \varphi \circ \iota_1 : U_1 \to V$ and $\varphi_2 = \varphi \circ \iota_2 : U_2 \to V$.
Thus the isomorphism you seek is
\begin{align*}
\hom_R(U_1, V) \oplus \hom_R(U_2,V) &\longrightarrow \hom_R(U_1 \oplus U_2, V)\\
(\varphi_1, \varphi_2) &\longmapsto \varphi_1 \oplus \varphi_2\\
\end{align*}
with inverse
\begin{align*}
\hom_R(U_1 \oplus U_2, V) &\longrightarrow \hom_R(U_1, V) \oplus \hom_R(U_2,V)\\
\varphi &\longmapsto (\varphi \circ \iota_1, \varphi \circ \iota_2) \, .
\end{align*}
