Does $[G,G] \trianglelefteq \text{ker}(\Psi)$ hold?

By definition, $[G,G]$ is generated by elements of the form $[x,y]=xyx^{-1}y^{-1}$ (and not $xy^{-1}$). It suffices to show that every commutator is in $\ker \psi$ to show that $[G,G]\subset \ker \psi$. But

\begin{align} \psi(xyx^{-1}y^{-1}) &= \psi(x)\psi(y)\psi(x^{-1})\psi(y^{-1})\\ &= \psi(x)\psi(x)^{-1}\psi(y)\psi(y)^{-1}\\ &= 1 \end{align} Because $H$ is abelian.

The fact that $[G,G]$ is normal is because every conjugate of a commutator is a commutator : \begin{align} z [x,y]z^{-1} &= zxyx^{-1}y^{-1}z^{-1}\\ &=zxz^{-1}zyz^{-1}(zxz^{-1})^{-1}(zyz^{-1})^{-1}\\ &= [zxz^{-1},zyz^{-1}] \end{align} Thus, if $g = [x_1,y_1]\cdot [x_2,y_2]\cdots[x_n,y_n] \in [G,G]$, and if $h \in G$ : \begin{align} hgh^{-1} &= h\left([x_1,y_1]\cdot [x_2,y_2]\cdots[x_n,y_n]\right)h^{-1}\\ &= h[x_1,y_1]h^{-1}h[x_2,y_2]h^{-1}\cdots h[x_n,y_n]h^{-1}\\ &= [hx_1h^{-1},hy_1h^{-1}]\cdot[hx_2h^{-1},hy_2h^{-1}]\cdots[hx_nh^{-1},hy_nh^{-1}] \in [G,G] \end{align}