Suppose $A$ is a commutative ring with unit, and $M$ is a finitely-generated module with the surjection $\pi: A^n \twoheadrightarrow M$.
Let $f : M \to M$ be a module homomorphism. I am trying to see if there exists a a module homomorphism $\hat{f}: A^n \to A^n$ such that $f \circ \pi = \pi \circ \hat{f}$.
For an element $a\in A^n$, I can compute $f \circ \pi(a)$ and lift to a representative $a' \in A^n$ and set $\hat{f}(a)=a'$. But is it true that I can always pick my representatives such that $\hat{f}$ ends up being a homomorphism?
You should not try assigning a representative to each $a \in A^n$ independently. Instead, let $\left(e_1,e_2,\ldots,e_n\right)$ be the standard basis of the $A$-module $A^n$. For each $i\in\left\{1,2,\ldots,n\right\}$, choose some $g_i \in A^n$ such that $f\left(\pi\left(e_i\right)\right) = \pi\left(g_i\right)$. Then, let $\widehat f : A^n \to A^n$ be the $A$-linear map which sends every $e_i$ to $g_i$. Then, the $A$-linear maps $f \circ \pi$ and $\pi \circ \widehat f$ are equal to each other on the basis $\left(e_1,e_2,\ldots,e_n\right)$ of $A^n$, and thus are identical.