The overall goal of the problem is the following:
Let $R$ be a local commutative ring (so that there is a unique maximal ideal $\mathfrak{m}$) and let $M$ be a finitely generated $R$-module. Assume there is an $R$-module $N$ such that $M \oplus N$ is a free $R$-module. Prove $M$ is free.
Assume that the following have already been proved:
(1) We may choose $m_1, \dots, m_r \in M$ such that $\left\{ m_\alpha + \mathfrak{m}M \right\}$ forms a basis for the vector space $M/\mathfrak{m}M$ over $R/\mathfrak{m}$, and $M = \left< m_1, \dots, m_r \right>$.
(2) Let $\left\{e_1, \dots, e_r\right\}$ be a basis for the free $R$-module $F := R^r$. Let $\pi \colon F \to M$ be the surjective $R$-module homomorphism defined by $e_\alpha \mapsto m_\alpha$. Then $\pi$ is split. In particular, there is an $R$-module isomorphism $\varphi \colon F \to \ker\pi \oplus M$.
Here is where I'm stuck:
Prove that $\ker\pi / \mathfrak{m}\ker\pi = 0$.
Thus by Nakayama's Lemma, $\ker\pi = 0$. Conclude $M$ is free.
Now, I have been trying to show $\ker\pi \subset \mathfrak{m}\ker\pi$ directly. Using (1), I was able to show that if $x = \sum r_\alpha e_\alpha \in \ker\pi$, then $r_\alpha \in \mathfrak{m}$ for all $\alpha$. But I cannot progress far from here. Notice that (2) should of course be used here; if we could prove $\ker \pi = 0$ without it, then we would still conclude $M$ is free. Also, note that (1) is not used to prove (2). So we really need to use both conditions to prove the needed result here. Perhaps, based on Aluffi's phrasing, it is best not to try to show $\ker\pi \subset \mathfrak{m}\ker\pi$ directly. I don't know.
Thanks everyone for your help.
You almost done it.
Consider $F=\ker\pi\oplus M$, $x\in\ker\pi$, $x=\sum r_ie_i$ with $r_i\in\mathfrak m$, and write $e_i=u_i+v_i$ with $u_i\in\ker\pi$ and $v_i\in M$. Then $x=\sum r_iu_i+\sum r_iv_i$, so $$x-\sum r_iu_i=\sum r_iv_i\in\ker\pi\cap M=\{0\},$$ hence $x=\sum r_iu_i\in\mathfrak m\ker\pi$.