Are there atoms of homomorphisms defined on a direct sum? Suppose that we have an $R$-module homomorphism
$$
f: M_1 \oplus M_2 \longrightarrow N_1 \oplus N_2, 
$$
where each $M_i$ and $N_i$ are $R$-modules.
For each $i=1, 2$, I want to find an $R$-module homomorphism
$$
f_i: M_i \longrightarrow N_i
$$
such that
$$
f((m_1,m_2)) = (f_1(m_1), f_2(m_2))
$$
for all $m_i \in M_i$.
My Question: Can I find such a homomorphism? Are such homomorphisms always available?
 A: No, in general they don’t exist. You need more information to reconstruct $f$.
For example, take $R$ to be $\mathbb{Z}$, $M_1=M_2=N_1=N_2=\mathbb{Z}_2$, the integers modulo $2$, and consider the map $f\colon\mathbb{Z}_2\oplus\mathbb{Z}_2\to\mathbb{Z}_2\oplus\mathbb{Z}_2$ given by $f(a,b) = (a,a)$. Since $M_2$ is the kernel of the map, any morphism $f_2\colon M_2\to N_2$ with $f(x,y) = (f_1(x),f_2(y))$ would require $f_2(y)=0$ (since $f(0,y) = (0,0)$ and $f(x,y) = f(x,0)+f(0,y)$). But the composition of $f$ with the projection $N_1\oplus N_2\to N_2$ is nontrivial, so this cannot be done.
What you can do is find four homomorphisms, $f_{ij}\colon M_i\to N_j$ such that
$$f(x,y) = (f_{11}(x)+f_{21}(y), f_{12}(x)+f_{21}(y)).$$
Namely, $f_{ij}$ is the composition of the embedding $\iota\colon M_i\hookrightarrow M_1\oplus M_2$, with $f$, with the projection $\pi_j\colon N_1\oplus N_2\to N_j$.
A: No, you can't in general.
Take $R$ an arbitrary ring and $N_1,N_2$ arbitrary $R$-modules. Let $M_1=N_1\oplus N_2$, and let $M_2=\{\star\}$ be the trivial module, and consider the obvious map $$f: M_1\oplus M_2\rightarrow N_1\oplus N_2: ((a,b), \star)\mapsto (a,b).$$ Unless $N_2$ is trivial, we can't find a decomposition of the type you're looking for.
A: Not necessarily, but composing with the canonical homomorphisms
\begin{cases}
M_i\hookrightarrow M_1\oplus M_2,&i=1,2 \\
N_1\oplus N_2\rightarrow N_i,&i=1,2
\end{cases}
you can deduce homomorphisms $u_{ij}:M_i\longrightarrow N_j\quad (i,j=1,2)$ such that the homomorphism $f$ is represented by the matrix $A$:
$$f(m_1,m_2)=\underbrace{\begin{pmatrix}u_{11}& u_{21}\\u_{12}&u_{22 }\end{pmatrix}}_A\begin{pmatrix}m_1\\m_2\end{pmatrix}.$$
