About how to understand the third isomorphism theorem

The following diagram is one classical way that the theorem is frequently "illustrated" (excerpted from Burris and Sankappanavar's A Course in Universal Algebra). Essentially the diagrams illustrate the set-theoretic view of the matter, in terms of equivalence classes or, equivalently, partitions. Before studying the algebraic versions of these theorems, for motivation, it helps to have a good grasp on their much simpler set-theoretic versions, i.e. "forget" the algebraic structure (operations), so reducing from algebras and their maps (homs), to simply sets and functions. Then a congruence reduces to an equivalence relation (no operations to preserve), i.e. a partition, which leads to the vivid depiction below, in terms of finer and coarser partitions.

A very silly (and not "mathematically" rigorous) way I remember the theorem is by examining the analogue in rational numbers. Everyone who knows anything gets that $\dfrac{a/b}{c/b} = \dfrac{a}{b}\cdot\dfrac{b}{c} =\dfrac{a}{c}$.

We can intuitively think of the Third Isomorphism theorem in the same fashion. $$\dfrac{A/B}{C/B} ``=" \dfrac{A}{B}\cdot\dfrac{B}{C} \cong A/C$$

Again, not mathematically valid, but a good mnemonic.

The best motivation I have comes from the classical groups: $\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$. In this way when you start with $m|n$ so that we have a subgroup relation to talk about

$$\mathbb{Z}/n\mathbb{Z}\bigg/n\mathbb{Z}/m\mathbb{Z}\cong \mathbb{Z}/(n/m)\mathbb{Z}$$

so that things really work like fractions, which is how the notation is setup, and which motivates the idea for more general groups.

This also works if you start with $A=\mathbb{Z}/n\mathbb{Z}$. Then taking $k|m|n$ we have

$$A/kA\bigg/mA/kA\cong A/(m/k)A$$

This is comparable to actual fractions since it looks like

$${n\over k}\bigg/{n\over m}={m\over k}$$