Proving the kernel is a normal subgroup.

The argument goes as:

Let $f: G \to G_1$ be the homomorphism, and let $H = \operatorname{Ker}(f) = \{g \in G \; | \; f(g)=e \}$ be its kernel.

A subgroup is normal if it is invariant by conjugation. So fix an arbitrary element $h \in H$. For all $g \in G$, is it true that $ghg^{-1} \in H$ (which would mean that $H$ is indeed invariant by conjugation)?

And then he proves it as you have done above: $$f(ghg^{-1}) = f(g)f(h)f(g^{-1}) = f(g)f(g)^{-1} = e$$

We have a group $G$ and a group homomorphism $f$ that takes elements from $G$. We also have $\ker(f) = H$. We want to show that $H$ is normal. One way of describing "$H$ is normal" is that given arbitrary $h\in H$ and $g\in G$, we must have $ghg^{-1}\in H$. By definition of $H$, $ghg^{-1}\in H$ is equivalent to $f(ghg^{-1}) = e$. So that is what we check for.