Let $R$ be a ring and let $M \to N \to K \to 0$ be an exact sequence of $R$-modules. Is there an exact sequence of free modules $A \to B \to C \to 0$ and a commutative diagram $$\begin{array}{c} M & \rightarrow & N & \rightarrow & K & \rightarrow & 0 \\ \uparrow & & \uparrow && \uparrow \\ A & \rightarrow & B & \rightarrow & C & \rightarrow & 0\end{array}$$ in which the vertical homomorphisms are epimorphisms?
I can find such a diagram in which $B \to N$ and $C \to K$ are epimorphisms: Choose any epimorphism $C \to K$ with $C$ free. Choose any epimorphism $B \to N \times_K C$ with $B$ free. Finally, choose any epimorphism $A \to \ker(B \to C) \times_N M$ with $A$ free. This works fine, but it is not clear if $A \to M$ is an epimorphism (probably not!).
The map $A\rightarrow M$ is an epimorphism in your construction, since the image of $\ker(B\rightarrow C)\rightarrow N$ is $\ker(N\rightarrow K)$, which is the image of $M\rightarrow N$.