I came across this passage in Rotman's Advanced Modern Algebra:
If $M$ is finitely presented, there is a short exact sequence
$\begin{equation*}0 \rightarrow K \rightarrow F \rightarrow M \rightarrow 0,\end{equation*}$
where $F$ is free and both $K$ and $F$ are finitely generated. Equivalently, $M$ is finitely presented if there is an exact sequence
$\begin{equation*}F' \rightarrow F \rightarrow M \rightarrow 0,\end{equation*}$
where both $F'$ and $F$ are finitely generated free modules (just map a finitely generated free module onto $K$).
It's not obvious to me why the second statement must be true. Is it really trivial enough to just claim in passing? Any insight would be much appreciated. Thank you.
Yes, it is equivalent. If $K$ is finitely generated then there is a generating set $k_1, \ldots, k_n$. Let $F'$ be free of rank $n$ and map $F' \to K$ by sending the generators of $F'$ to $k_1, \ldots, k_n$. Then the map $F' \to F$ is the composition $F' \to K \to F$.
Conversely if you have $\phi\colon F' \to F$ then define $K = F'/\ker\phi$. The module $K$ is finitely generated (use the generators of $F'$) and $\phi$ induces an injective map $K \to F$.
So basically the difference between a SES $0 \to K \to F \to M \to 0$ and a two step free resolution $F' \to F \to M \to 0$ is whether or not you've encoded the relations between the generators in the map $F' \to F$ or in the module $K$.