The isomorphism $(A\otimes _A M) / (\mathfrak{a} \otimes _A M) \cong M/ \mathfrak{a} M$ is neither true nor false!
Indeed, the left side does not make sense, since the $A$-module $\mathfrak{a} \otimes _A M$ is not a submodule of $A\otimes _A M$. Let's analyze this.
Start with the exact sequence $0\to \mathfrak{a}\to A\to A/ \mathfrak{a} \to 0$.
Tensoring it with $M$ you get the exact sequence $\mathfrak{a}\otimes_A M \to A \otimes _A M \to A/ \mathfrak{a}\otimes_A M \to 0$ or
$\mathfrak{a}\otimes_A M \stackrel {f}{\to} M \to M/ \mathfrak{a} M \to 0$ so that you get the correct isomorphism
$M/f(\mathfrak{a}\otimes_A M)\simeq M/ \mathfrak{a} M$
The point to keep in mind is that $f:\mathfrak{a}\otimes_A M \stackrel {f}{\to} M:\alpha\otimes m\mapsto \alpha.m$ is not injective in general (unless you know that $M$ is flat over $A$) and it is thus false to consider that $\mathfrak{a}\otimes_A M$ is a submodule of $M$ (or of $A\otimes_AM)$.
An explicit counterexample to the injectivity of $f$
Let $p\in \mathbb Z$ be a prime, $A=\mathbb Z/(p^2)$, $M=\mathbb Z/(p)$ and $\mathfrak{a}=pA$.
The morphism $f:\mathfrak{a}\otimes _A M\to M:pa\otimes m\mapsto pam=0$ is zero hence $f$ is not injective because $\mathfrak{a}\otimes _A M=\mathfrak{a}\otimes _{A/p} M\neq 0$ .
[The last equality is due to the fact that $p$ kills both $\mathfrak{a}$ and $M$, so that the tensor products over $A$ or over $A/p$ are the same. The inequality is due to the fact that $A/p=\mathbb F_p$, a field: over a field the tensor product of two nonzero vector spaces is nonzero.]