Note: $M,M',M''$ are modules.
In order to show that it's split exact, I understand that I need to show that $\beta:M\rightarrow M''$ has a left inverse, and similarly that $\alpha:M'\rightarrow M$ has a right inverse. I also understand that $M''$ must have a generating set (as it's free).
$\alpha$ is a monomorphism (as the sequence is exact), so I'd imagine that as it's 1-1 it must have an inverse. I'm not sure about what to do with $\beta$ though, as I only know it's an epimorphism.
EDIT: here I'm writing my function as $x(f)$ rather than $f(x)$
(answer from a discussion with @peter a g)
Our objective is to form a homomorphism from $M''$ to $M$.
Call $\{e_1,e_2,...,e_n\}$ a basis of the free module $M''$. As it is a basis, every element of $M''$ can be written as a linear combination of the basis elements in a unique way.
At this point, we construct our homomorphism from $M''$ to $M$ by first choosing the elements $a_i$ of $M$ that map to the elements of the basis $e_i$ under $\beta$, then defining our homomorphism by mapping each element $e_i$ of the basis to an $a_i$ and then extending this map so the domain is the whole of $M''$.