Let $A$ be a commutative ring. Let $M, N$ be $A$-modules and let $I\subset A$ be an ideal contained in the annihilators of $M$ and $N$, respectively. That is, $I\subset\mathrm{Ann}(M)\cap\mathrm{Ann}(N)$. Assume further that $f:M\to N$ is a $A$-linear map.
Both modules $M$ and $N$ have a canonical $A/I$-module structure given by $\bar{a}\cdot x=ax$ for every $\bar a\in A/I$, any representative $a\in\bar a$ and every $x\in M, N$. Does the $A$-linear map $f$ descend to a $A/I$-linear map via the rule
$$f(\bar a\cdot x)=f(ax)=af(x)=\bar a\cdot f(x)?$$
To me the above fact is evident but on the other hand it cannot be found in any textbook.
The observation is correct. My opinion regarding the unavailability of this as a theorem is perhaps because it can be regarded as a particular case of the following:
If $M$ and $N$ are $A$-modules, $B$ an $A$-algebra, then any $A$-module map $f:M\to N$ induces a $B$-module map $f\otimes 1: M\otimes B\to N\otimes B$ such that $f(m\otimes b)=f(n)\otimes b$ for all $m\in M,n\in N,b\in B$.
In the present case, we can take $B=A/I$. Then $M\otimes (A/I)\cong M/IM\cong M$ as $A/I$-modules and a similar thing for $N$. Then $f\otimes 1$ as an $A/I$-module map from $M$ to $N$ will look like the the map that is described in the question.