I just proved a result, that a short exact sequence with a free module at the third position is split exact, i.e., the short exact sequence $$ 0 \to \mathbf{M}' \to \mathbf{M} \to \mathbf{M}'' \to 0 $$ splits if $\mathbf M''$ is a free $R$-module.
I was wondering would the converse of this be true?
No. For any pair of modules $A, B$ we always have the short exact sequence:
$0 \rightarrow A \hookrightarrow A \oplus B \twoheadrightarrow B \rightarrow 0$
where the arrows are the obvious ones. There is no reason to expect $A \oplus B$ to be free in general.
For example, you can take $A$ and $B$ to be the $C^\infty(M)$-modules of global sections of some pair of non-trivial vector bundles on a smooth manifold $M$. Even more concretely, take $A=B=\mathcal{T}(S^2)$ to be the module of smooth vector fields on the $2$-sphere.