Let $\varphi:G\rightarrow H$ be a homomorphism of groups. I'm trying to understand why the induced map $\varphi'(xG')=\varphi(x)H'$ is well defined.
So if $xG'=yG'$ then $xy^{-1}\in G'$ and then $\varphi'(xy^{-1}G')=\varphi(xy^{-1})H'=\varphi(x)\varphi(y)^{-1}H'$ and also $\varphi'(xy^{-1}G)=\varphi'(G)=\varphi(1)H'=H'$, and so $\varphi(x)H'=\varphi(y)H'$.
Is this enough to show well defined? There was no use of any properties of the commutator.
I didn't understand your reasoning, but in general if $G_0 \subset G$ and $H_0 \subset H$ are normal subgroups, and $\varphi: G \rightarrow H$ is a homomorphism mapping $G_0$ into $H_0$, then the homomorphism $\overline{\varphi}: G/G_0 \rightarrow H/H_0$ sending $gG_0$ to $\varphi(g) H_0$ is well defined.
In fact, it suffices to show that if $S$ is a subset of $G$ which generates $G_0$, then $\varphi(S) \subset H_0$.
Granting the above, all you need to show in your case is that commutators in $G$ are sent to commutators in $H$ by $\varphi$.