Homomorphism of local rings

It is clear that $\phi(\mathfrak{m}) + \mathfrak{n}^2 \subset \mathfrak{n}$. Take an element $a \in \mathfrak{n}$. Since $\mathfrak{m} \rightarrow \mathfrak{n}/\mathfrak{n}^2$ is surjective, there exists $x \in \mathfrak{m}$ such that $a - \phi(x) \in \mathfrak{n}^2$. Thus, $a = \phi(x) + b$ for some $b \in \mathfrak{n}^2$. This shows that $\mathfrak{n} \subset \phi(\mathfrak{m}) + \mathfrak{n}^2$, and we are done.

I hope I am not misunderstanding what the map $\mathfrak{m} \rightarrow \mathfrak{n}/\mathfrak{n}^2$ is. I also want to say that you did all the difficult parts already.

Also take a look at Exercise 10 from Atiyah-Macdonald Chapter 2. While your proof using Snake Lemma is very nice, one can prove this statement by using Nakayama twice. The first time you use Nakayama is to deduce from $\phi(\mathfrak{m}) + \mathfrak{n}^2 = \mathfrak{n}$ that $\phi(\mathfrak{m}) = \mathfrak{n}$. This gives you that $\mathfrak{n} = \mathfrak{m}B$, i.e., the extension of $\mathfrak{m}$ in $B$ is $\mathfrak{n}$. Now, we know that the map $A\mathfrak{m} \rightarrow B/\mathfrak{m}B$ is surjective. Hence, by an argument similar to the one I gave above, you can show that $im(\phi) + \mathfrak{m}B = B$. Since $B$ is a finitely generated $A$-module, you use Nakayama again to deduce that $im(\phi) = B$.

This also shows you that you do not need $A$ Noetherian, and it suffices to only assume that the maximal ideal of $B$ is finitely generated.