I am looking for a residually finite semidirect product with the following properties. This is related to this question: If $K\rtimes \mathbb{Z}$ is a finitely generated group but $K$ isn't, can the following abelianization be finite? (Thank you, Derek Holt and Ycor for the answer to my previous question).
Suppose that we have a finitely generated and residually finite group $G = K\rtimes\mathbb{Z}$, but $K$ is not finitely generated. Let $T$ be a finite subset of $K$ such that $\langle (0,1), (k,0)\mid k \in T \rangle$ is a generating set of $G$. Let $\phi(n)$ be the automorphism on $K$ corresponding to $n \in \mathbb{Z}$, which defines the semi-direct product. Let $H = \langle T \rangle$, a finitely generated subgroup of $K$.
Now, suppose in addition, the automorphisms satisfy that $$\ldots \phi(-1)(H) \supsetneqq \phi(0)(H) \supsetneqq \phi(1)(H) \supsetneqq \phi(2)(H) \ldots \tag{$*$}$$
Because $G$ is residually finite, $H$ is, too. we know that there exists a sequence of nested, normal, finite index subgroups $$H = N_0 \rhd N_1 \rhd N_2 \ldots $$ with the trivial intersection.
Question: Is it possible that all $N_i$ have finite abelianization, i.e. the quotient group $N_i \big/ [N_i,N_i]$ is finite for every $i$?
My thoughts so far: Using the answer from the previous question, here is a group that almost satisfies all the properties: let $H$ be the Grigorchuk group, and we construct $K$ using the ascending HNN-extension with an injective non-surjective endomorphism of $H$. Unfortunately, such $K$ is not residually finite (in general, requiring that the HNN is residually finite is a strong condition), but $K\rtimes \mathbb{Z}$ satisfies all the other properties.
I think the derive series $H \supset H^{(1)}\supset H^{(2)} \dots $ of $H$ never stablises, and each $H^{(i)}$ is finite index in $H^{(i+1)}$. Also, we can see that $H \rhd [N_0,N_0] \rhd [N_1,N_1] \rhd [N_2,N_2] \ldots $ is another nested, normal, finite index subgroups of $H$ with trivial intersection.
Any help would be really appreciated.