I offer a solution that is very similar to the one given by Ben, but I cite specific results used from Chapter 2 of Atiyah-Macdonald and give a slightly different presentation. I leave this here only because it may be useful to some readers who are struggling with this chapter, like I was when I first worked through it.
Problem:
Let $A$ be a commutative ring with unity, $\mathfrak{a}$ an ideal, $M$ an $A$-module. Show that $(A/\mathfrak{a}) \otimes_A M \cong M/\mathfrak{a}M.$
Solution:
Proceeding by way of the hint, consider the exact sequence $$0 \to
\mathfrak{a} \xrightarrow{\pi} A \xrightarrow{\phi} A/\mathfrak{a} \to
0.$$
The map $\pi$ is the standard inclusion, $\pi(a) = a,$ for all $a \in
\mathfrak{a}$. It is easy to see that $\pi$ is injective and
$\mathrm{Im}(\pi) = \mathfrak{a}$. The map $\phi$ is the standard
surjection, $\phi(a) = a + \mathfrak{a},$ for all $a \in A$. It is
easy to see that $\mathrm{Ker}(\phi) = \mathfrak{a}$. Since
$\mathrm{Im}(\pi) = \mathrm{Ker}(\phi)$, $\pi$ is injective, and
$\phi$ is surjective, we are satisfied that this is in fact a short
exact sequence.
Since each object in the short exact sequence can be seen as an
$A$-module, we can tensor with the $A$-module $M$, which by
$\textbf{Proposition 2.18}$ (page 28) induces a new exact sequence $$
\mathfrak{a} \otimes_A M \xrightarrow{\pi \, \otimes \, 1} A \otimes_A
M \xrightarrow{\phi \, \otimes \, 1} A/\mathfrak{a} \otimes_A M \to
0.$$
Referring to $\textbf{Proposition 2.14, iv)}$ (page 26) we know we
have an isomorphism \begin{align}\varphi &\colon A \otimes_A M \to M\\
\varphi &\colon a \otimes x \mapsto ax. \end{align}
Since our goal is to make a claim $M/\mathfrak{a}M$ is isomorphic to
$A/\mathfrak{a} \otimes_A M$ it is a natural thought, with the first
isomorphism theorem in mind, to try and use what we have so far to
cook up a surjective map from $M$ to $A/\mathfrak{a} \otimes_A M$, and
hope the kernel is $\mathfrak{a}M$. Thus the natural candidate is to
consider $$(\phi \otimes 1) \circ \varphi^{-1} \colon M \to
A/\mathfrak{a} \otimes_A M.$$
First see that by exactness of the tensored sequence $\phi \otimes 1$
must be a surjective morphism. Since $\varphi$ is an isomorphism,
$\varphi^{-1}$ must also be a surjective morphism, so $(\phi \otimes
1) \circ \varphi^{-1}$ is a composition of surjective morphisms and
thus is also a surjective morphism. But we also have
\begin{align*} \mathrm{Ker}\left((\phi \otimes 1) \circ
\varphi^{-1}\right) &= \varphi \left( \mathrm{Ker}(\phi \otimes 1)
\right) \tag{prove this if it is not obvious why}\\ &=
\varphi(\mathfrak{a} \otimes_A M) \tag{$\mathrm{Im}(\pi \otimes 1) = \mathrm{Ker}(\phi \otimes 1)$}\\ &=\mathfrak{a}M \tag{definition of $\varphi$}. \end{align*}
Thus by the first isomorphism theorem for modules, we conclude that
$(A/\mathfrak{a}) \otimes_A M \cong M/\mathfrak{a}M.$