Prove that the following quotients are isomorphic I am preparing myself for an upcoming exam, and I've found the following problem

Let $M$ be a $R$-module and $N_1 \subset N_2 \subset M$ be $A$-submodules. Use the Snake Lemma to show that $$\frac{M/N_1}{N_2/N_1} \cong \frac{M}{N_2}.$$

I have tried several different setups to use the Snake Lemma and prove this result. However, none of them seem to work. Could anyone please point out how I should start?

Edit: The Snake Lemma states that, in an abelian category, given the following commutative diagram
$\require{AMScd}$
\begin{CD}
@. A @>{f}>> B @>{g}>> C @> >> 0\\
@. @V{a}VV @V{b}VV @V{c}VV\\
0 @>>> A' @>{f'}>> B' @>{g'}>> C'
\end{CD}
in which the rows are exact sequences and $0$ is the zero object, one has the following exact sequence
$$\operatorname{Ker}(a) \rightarrow \operatorname{Ker}(b) \rightarrow \operatorname{Ker}(c) \xrightarrow{d} \operatorname{Coker}(a) \rightarrow \operatorname{Coker}(b) \rightarrow \operatorname{Coker}(c)$$
with $d$ a homomorphism.
 A: Construct the following commutative diagram:
$\require{AMScd}$
\begin{CD}
0@>>> N_2 @>>> M @>>> \frac M{N_2} @>>> 0\\
@. @VVV @VVV @VVV\\
0 @>>> \frac{N_2}{N_1} @>>> \frac M{N_1} @>>> \frac{M/N_1}{N_2/N_1} @>>>0
\end{CD}
The rows are the usual exact sequence associated to a quotient modules, the first two vertical arrows are canonical and commutativity as well as the existence of the latter vertical arrow follow from the homomorphism theorem. Let $\varphi$ be the last vertical arrow.
Now apply the Snake lemma, or rather as slightly upgraded version thereof. As you can see we have short exact sequences rather than only right and left, respectively, exact rows. One can show (quite easily) that in this case the Snake Lemma is sequence is exact at $\ker a$ and $\operatorname{coker} c$ too. Using this we see that there are exact sequences $0\to N_1\to N_1\to\ker\varphi\to0$ and $0\to\operatorname{coker}\varphi\to0$ implying that $\ker\varphi=0=\operatorname{coker}\varphi$. This in turn is equivalent to $\varphi$ being an isomorphism as desired.
