So far I have showed that $K$ is a normal subgroup and that the Operation defined on $G/K$ is abelian.
Now I have to show that if $\phi:G\rightarrow A$ is a group
homomorphism and $A$ is abelian. Then there exists a
$\phi':G/K\rightarrow A$ such that $\phi$ can also be expressed as
$\phi '\circ\pi$, where $\pi$ is the natural homomorphism
$\pi:G\rightarrow G/K$.
I know since $\phi$ is a homomorphism that $G/\ker(\phi)\cong \text{Im}(\phi)$.
So there exists a bijection $f:G/\ker(\phi)\rightarrow \text{Im}(\phi)$.
But I don't know how to prove that $\ker(\phi)=K$.
What I do know is that
$$z\in\{xyx^{-1}y^{-1},x\wedge y\in G\}=:S\Longrightarrow z\in \ker(\phi)\Longrightarrow S\subseteq\ker(\phi)$$
Also by Definition of $K$, I know that $S\subseteq K$.
Now there might be elements in $K$ which cannot be expressed this way. Also there might be elements in the kernel which are not in $K$.
Best Answer
If $K$ and $N$ are normal subgroups and $K\leq N$ then the map $G/K\to G/N$ prescribed by $gK\mapsto gN$ is well defined and is a group homomorphism.
Denoting this map by $\nu$ for the natural homomorphisms $\pi_N:G\to G/N$ and $\pi_K:G\to G/K$ we find:$$\pi_N=\nu\circ\pi_K\tag1$$
Now if $\phi:G\to A$ is a group homomorphism where $A$ is abelian then for the subgroup $K$ mentioned in your question we find: $K\leq N:=\mathsf{ker}\phi$.
As usual we can write $\phi=\psi\circ\pi_N$ for a group homomorphism $\psi:G/N\to A$ and applying $(1)$ we find:$$\phi=\psi\circ\nu\circ\pi_K=\phi'\circ\pi_K$$
Here $\phi'=\psi\circ\nu:G/K\to A$ is a composition of group homomorphisms, hence is a group homomorphism.