Let $G$ be the semi-direct product of a group $A$ by a group $B$ where $B$ is normal in $G$. Furthermore, let $N = [A,B]$ denote the subgroup of $G$ generated by all commutators $[a, b] = aba^{-1}b^{-1}$, where $a \in A$ and $b \in B.$
I am trying to show that $G/N \cong A \times (B / N)$.
Here is my thought process: First I show that $N$ is normal in $G$. ( This is straight forward and has been done). Secondly I define a mapping $$\varphi: G \rightarrow A \times (B / N)$$ $$g= ab \mapsto (a, b[a,b])$$ I will need to show that $\varphi$ is a surjective homomorphism whose kernel is $G'$. Then by the first isomorphism theorem I can conclude that $G/N \cong A \times (B / N)$.
My problem is showing that $\varphi$ is a homomorphism. That is to say for any $g_1, g_2 \in G$ such that $g_1 = ab$ , $g_2 = cd$ where $a,c \in A$, and $b,d \in B$, we have that $\varphi(g_1 \cdot g_2) = \varphi(g_1) \cdot \varphi(g_2)$.
Is my mapping sufficient? Is my thought process correct? Any help would be appreciated.
More generally, suppose that $G=K\rtimes H$ (here $K$ is the normal subgroup, $H$ acts on it), and that $N$ is a subgroup of $K$ that is normal in $K$, and that is $H$-invariant; that is, for all $n\in N$ and $h\in H$, $hnh^{-1}\in N$. We want to show that $N$ is normal in $G$, and that $G/N\cong (K/N)\rtimes H$, with the action of $H$ on $K/N$ being the one induced by the action of $H$ in $K$.
Though I am thinking of $G$ as an internal semidirect product, I will use the ordered pair notation: so elements of $G$ are written uniquely as $(h,k)$ with $h\in H$, $k\in K$, and the product is given by $$(h,k)*(h',k') = (hh',k^{h'}k') = (hh',(h')^{-1}kh'kk').$$
Indeed, since $G$ is generated by $K$ and $H$, and $N$ is invariant under conjugation by elements of both $K$ and $H$, it follows that $N\triangleleft G$. Note also that the action of $H$ on $K$ induces an action of $H$ on $K/N$ by conjugation, because $N$ is mapped to itself.
To show that $G/N\cong (K/N)\rtimes H$, we take the map defined (as you kind of almost did) by $\varphi\colon G\to (K/N)\rtimes H$ by $\varphi(h,k) = (h,kN)$.
This is a homomorphism: given $(h_1,k_1)$, $(h_2,k_2)$ in $G$, we have $$\begin{align*} \varphi(h_1,k_1)\varphi(h_2,k_2) &= (h_1,k_1N)(h_2,k_2N)\\ &= (h_1h_2, (k_1N)^{h_2}k_2N) \\ &= (h_1h_2,(k_1^{h_2}Nk_2N)\\ &= (h_1h_2,k_1^{h_2}k_2N);\\ &=\varphi(h_1h_2,k_1^{h_2}k_2)\\ &= \varphi\bigl( (h_1,k_1)(h_2,k_2)\bigr). \end{align*}$$ The kernel of $\varphi$ consists of the $(h,k)\in G$ such that $\varphi(h,k)=(h,kN)=(e,eN)$. Thus, we must have $h=e$ and $k\in N$. Conversely, all such elements are in the kernel.
Finally, $\varphi$ is surjective, since the element $(h,kN)$ of $(K/N)\rtimes H$ is the image of $(h,k)\in G$. Thus, we have that $$\frac{G}{N} \cong \frac{K}{N}\rtimes H.$$
Now, in your specific situation, $N=[K,H]$. This is a subgroup of $K$, since the generators all lie in $K$: $[k,h] = k(hk^{-1}h^{-1})\in K$.
It is invariant under the action of $H$, since given a generator $[k,h]$ and $y\in H$, we have $y[k,h]y^{-1}=[yky^{-1},yhy^{-1}]\in [K,H]$ (since $yky^{-1}\in K$).
It is normal in $K$ (see, e.g., here, though that uses a different commutator convention), since given $x\in K$ and a generator $[k,h]=khk^{-1}h^{-1}$ of $[K,H]$, we have $$\begin{align*} x[k,h]x^{-1} &= xkhk^{-1}h^{-1}x^{-1}\\ &= (xkhk^{-1})(x^{-1}h^{-1}hx)h^{-1}x^{-1}\\ &= \bigl((xk)h(xk)^{-1}h^{-1}\bigr) \bigl(hxh^{-1}x^{-1}\bigr)\\ &= [xk,h][h,x]\\ &= [xk,h][x,h]^{-1}. \end{align*}$$ This element lies in $[K,H]$, since it is the product of a generator and the inverse of a generator of $[K,H]$.
Thus, in your situation we have $G/[K,H]\cong (K/[K,H])\rtimes H$.
It only remains to show that this semidirect product is actually a direct product. For that, we just need to show that the action of $H$ on $K/[K,H]$ is trivial. Indeed, letting $N=[K,H]$, let $h\in H$ and $k\in K$. Then $$(kN)^h = h^{-1}khN = kk^{-1}h^{-1}khN = k([k^{-1},h^{-1}])N = kN,$$ since $[k^{-1},h^{-1}]\in N$. Thus, this is a semidirect product with a trivial action, and so isomorphic to the direct product, as desired.