Taking quotient groups is, intuitively, "modding things out," i.e. setting things equal to zero and then living with the consequences (if $2=0$ then $1+1=0$ and $4=0$ and $1=3$ etc. for example).
This is why the congruence relation approach to quotient groups is discussed. Say $S$ is some subset of some group $G$. If you set everything in $S$ equal to the identity of $G$, there will be consequences to this. In particular, the product and inverses of anything in $S$ must also be the identity (since performing multiplication and inversion with just the identity element will yield nothing except for the identity element). Thus, if you mod out by $S$, you end up also modding out the subgroup that $S$ generates (the subgroup it generates is precisely the elements formed out of repeated multiplication and inversion of the elements of $S$). Moreover, conjugating an element of $S$ by anything from $G$ will also yield the identity, since conjugating the identity element yields the identity element. We end up not just modding out by $\langle S\rangle$, but by the conjugates of all of the elements therein, and then all of their products and inverses, and so on. Therefore, modding out by $S$ is the same as modding out by the normal subgroup generated by $S$.
In my opinion, this is an extremely important intuition in group theory.
Modding $A_1\times\cdots\times A_n$ out by $B_1\times\cdots\times B_n$ amounts to setting all elements of $B_1$ equal to the identity in the first coordinate $A_1$, etc. etc., and setting all elements of $B_n$ equal to the identity in the last coordinate $A_n$, all independently since one can form elements in $B_1\times\cdots B_n$ by choosing coordinates from the relevant $B_k$ all independently. This independence tells us that the product of quotients $A_1/B_1\times\cdots A_n/B_n$ achieves the exact same effect: in the $k$th coordinate we have the elements of $A_k$ except the elements of $B_k$ have been set equal to the identity. Since the whole setup is the same throughout (still have elements of $A_k$ for $k=1,\cdots,n$ put into coordinates), only things are being modded out, it's clear what the isomorphism $(\prod A_i)/(\prod B_k)\to\prod A_i/B_i$ should be: we simply take the residue $(a_1,\cdots,a_n)\bmod B_1\times\cdots B_n$ to $(a_1\bmod B_1,\cdots, a_n\bmod B_n)$.