Let $\varphi$ be a homomorphism of $G$ onto $G$ with ${\rm Ker} K$. Suppose $N$ is a normal subgroup of $G$ and let $N = \{x \in G | \varphi(x)\in N\}$.
Need to prove that $G/N$ is isomorphic to $(G/K)/(N/K)$.
This is what I have for now:
For $a, b\in G$, $$\sigma\varphi(ab) = \sigma(\varphi(ab)) = \sigma(\varphi(a)\varphi(b)) = \sigma(\varphi(a))\sigma(\varphi(b)) = \sigma\varphi(a)\sigma\varphi(b).$$
So $\sigma\varphi$ is a homomorphism.
If $a\in \text{Ker} \varphi$, then $\sigma\varphi(a) = \sigma(eH) = eK$ where $eH$ (resp. eK) is the identity of H (resp. $K$).
Therefore $a\in \text{Ker} \sigma\phi$. This implies that $\text{Ker} \phi \leqslant \text{Ker} \sigma\phi$. Furthermore, $\text{Ker} \phi / \text{Ker} \sigma\phi$.
Indeed, $\text{Ker} \phi / G$ so $\forall g\in \text{Ker} \sigma\phi\leqslant G, g \text{Ker} \sigma g−1 \subset \text{Ker} \phi.$
Moreover, if $\varphi$ and $\sigma$ are onto and $G$ is finite, then from the first isomorphism theorem,
$$|G| = | \text{Ker} \varphi||\varphi(G)| = | \text{Ker} \varphi||H|$$
and
$$|G| = | \text{Ker} \sigma\varphi||\sigma\varphi(G)| = | \text{Ker} \sigma\varphi||K|.$$
So $$ [\text{Ker} \sigma \varphi : \text{Ker} \varphi] = | \text{Ker} \sigma \varphi| /| \text{Ker} \varphi| = |G|/|K|/|G|/|H| = |H|/|K|.$$