Let $M$ be a finitely generated $R$-module. We need to show that there exists free R-modules $F_1, F_2$ of finite rank such that
\begin{equation}
F_1 \rightarrow F_2 \rightarrow M \rightarrow 0
\end{equation}
is an exact sequence.
There now exists a surjective homomorphism $\varphi \colon R^n \to M$ for some $n \geq 1$ such that $R^n/\ker \varphi \cong M$. Because $R$ is Noetherian, $R^n$ is also Noetherian and because $\ker \varphi$ is an ideal, we know that it is finitely generated. If I can now conlude that $\ker \varphi$ is free, then I have found a short exact sequence but I have no idea if this is even true.
Best Answer
In your post you produced an exact sequence
$$0 \rightarrow \ker(\phi) \rightarrow R^n \rightarrow M \rightarrow 0$$
and noted that $\ker(\phi)$ is finitely generated because it is an $R$-submodule of the Noetherian module $R^n$.
In general, if $$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$ is an exact sequence and $A' \rightarrow A$ is a surjection, then
$$A' \rightarrow B \rightarrow C \rightarrow 0$$ is exact too (where the first arrow is now the composition $A' \rightarrow A \rightarrow B$).
Can you finish it from there?