What is meant by: "the universal property of the canonical projection $p:G \to G/N$", is that if we have $f \in \mathrm{Hom}(G,G')$ such that $f(N) = \{e_{G'}\}$, then there exists a unique $f'$ with $f = f'\circ p$. This is just the First Isomorphism Theorem in disguise, the $f'$ in question is: $f'(gN) = f(g)$, which is well-defined since $f(n) = e_{G'}$ for all $n \in N$ (so if $gN = g'N$ then $g' = gn$ for some $n \in N$, thus $f(g') = f(gn) = f(g)f(n) = f(g)e_{G'} = f(g)$). This is often paraphrased as "$f$ factors through $p$".
So the "normal way" of showing $G/N \cong (G/M)/(N/M)$ is to show that $f(gN) = (gM)(N/M)$ is a well-defined isomorphism. But let's look at this another way:
The map $p_N: G \to G/N$ is a group morphism that kills $M$ (since $M \subset N$), so $p_N$ factors through the map $p_M: G \to G/M$ so that $p_N = f \circ p_M$, for some morphism $f$. Note that this morphism $f$ goes from $G/M$ to $G/N$ and kills $N/M$, so it in turn factors through $p_{N/M}:G/M \to (G/M)/(N/M)$, that is $f = f' \circ p_{N/M}$. But we also have the map $p_{N/M} \circ p_M$, which kills $N$, so there is a morphism $k$ such that $p_{N/M} \circ p_M = k \circ p_N$.
So $k \circ f' \circ p_{N/M} \circ p_M = k \circ f \circ p_M = k \circ p_N = p_{N/M} \circ p_M$, that is: $k \circ f' = \mathrm{id}_{(G/M)/(N/M)}$.
Similarly, $f' \circ k \circ p_N = f' \circ p_{N/M} \circ p_M = f \circ p_M = p_N$, so that $f' \circ k = \mathrm{id}_{G/N}$ (using implicitly the fact that the projections are epimorphisms to justify the cancellation).
These two facts together imply that $k$ and $f'$ are inverses, and thus isomorphisms. The important thing about all of this, is that we never mentioned any of the elements of $G$, or even any cosets, just homomorphisms between various groups.
Best Answer
actually there are a lot of examples, sometimes working with $\frac{H}{H\cap N}$ doesn't give the answer so you'll be obliged to use the 2nd thm of isomorphism. i'll give you an exmple whene you want to prove that a sub group of a solvable group is also solvable you'll take your quotients as $\frac{H}{H\cap G_{i}}$ where $G_{i}$ are the termes of the abelian serie of G