I often see if written that $$(R/M)/N = R/(M,N)$$ in reference to rings and modules (especially polynomial rings). I can see that it's probably related to the Isomorphism Theorems but I can find a rigorous proof that this holds in general.
Note, R is a ring and M,N are ideals.
A concrete example would be that $(\mathbb{R}[X]/(X))/(2) = \mathbb{R}[X]/(2,X)$
On
Note the quotient is not taken modulo $N$ (which is neither an ideal nor an $R$-submodule of $R/M$), but modulo the ideal generated by $N$ in $R/M$. Now this ideal is $$N\cdot R/M=(N+M)/M,$$ so the quotient you mention is but $$(R/M)\big/(N\cdot R/M)=(R/M)\big/(N+M)/M\simeq R/(N+M)$$ by the 3rd isomorphism theorem.
On
This is partly obscured by practical notation. If $R$ is a commutative ring and $I$ is an ideal:
So, you have to interpret the notation based on the context. To make matters more clear, I will add an overline for these reinterpretations; i.e. $\overline{r} = r + I$ and $\overline{J} = (I+J)/I$.
One reason why this notation is practical to use is that a lot of computations in this notation give the same result whether you compute them in $R$ or in $R/I$. For example:
Lemma: If $J$ is generated by a family of elements $\{ j_n \}$, then $\overline{J}$ is generated by $\{ \overline{j_n} \}$
A more precise formulation of the isomorphism you're looking at is
$$ (R/I) / \overline{J} \cong R / (I+J) $$
When you write out what $\overline{J}$ means, you'll see this is precisely one of the homomorphism theorems:
$$ (R/I) / ((I+J)/I) \cong R / (I+J) $$
Note that as stated, it technically doesn't make sense, as $N$ isn't an ideal of $R/M$. The correct formulation is $(R/M)/(N/M)\cong R/(M,N)$. This is more or less exactly the third isomorphism theorem.
Note that using this for exactly things like $$ (\mathbb{Z}[X]/(X))/(2) = \mathbb{Z}[X]/(2,X) $$ (i.e. modding out by one generator at a time, or changing the order of modding) is very common, but the $2$'s on either side are elements of different rings.