Let $M$, $N$, $P$ be $R$-module and functions $f:M\rightarrow N$, $g:M\rightarrow N$, $h:N\rightarrow P$, $k:P\rightarrow M$, which need not be homomorphisms. Define $f+g:M\rightarrow N$ by $(f+g)(a)=f(a)+g(a)$. It is easy to see that to have
$$h\circ(f+g)=h\circ f+h\circ g$$
$h$ needs to be a homomorphism. However, to have
$$(f+g)\circ k=f\circ k+g\circ k$$
why $k$ needs to be a homomorphism? It looks like the equality holds for arbitrary $k$.
$\def\ZZ{\mathbb{Z}}$No, it need not be a homomorphism.
For example, suppose that $R=\mathbb Z$, let $M=N=P=\ZZ/2\ZZ$, the abelian group of order two, let the two morphisms $f$, $g:\ZZ/2\ZZ\to\ZZ/2\ZZ$ be the identity map of $\ZZ/2\ZZ$ and let $k:\ZZ/2\ZZ\to\ZZ/2\ZZ$ be the constant map with value $1$. Then $(f+g)\circ k=f\circ k+g\circ k$, but $k$ is not a homomorphism.