Let $G$ be a second countable profinite group. Then $G$ can be written as a denumerable projective limit $\varprojlim_{i}(\cdots \to G_i\to G_{i-1}\to \cdots )$, where the $G_i$'s are finite and the connecting homomorphisms are surjective. Let $H$ be a finite group and let $G\rtimes H$ be a semidirect product of $G$ and $H$. My question is the following: Can $G$ be written as a denumerable projective limit $\varprojlim_{i}(\cdots \to G'_i\to G'_{i-1}\to \cdots )$ where the $G'_i$'s are finite and the connecting homomorphisms are surjective such that $$G\rtimes H= \varprojlim_{i}(\cdots \to G'_i\rtimes H\to G'_{i-1}\rtimes H\to \cdots )$$ where the connecting homomorphism $ G'_i\rtimes H\to G'_{i-1}\rtimes H$ are induced by $G'_{i}\to G'_{i-1}$?
It seems that the point is to find some open normal subgroups $N_i$ of $G$ such that they are $H$-stable, and then consider the finite quotients $G/N_i$.
The key step, as you pointed out, is how to construct an $H$-stable open normal subgroup of $G$ starting from a generic $N\unlhd_o G$. What follows should be what you were looking for.
Take $N\unlhd_o G$ and note that $N\leq_oG\rtimes H$ since its index is equal to $$|G\rtimes H : N|=|G/N||H|$$ that is finite. $N$ could not be $H$-stable, that is $N$ could not be a normal subgroup of $G\rtimes H$. To fix this problem, consider $$\tilde{N}:=\bigcap_{h\in H}N^h.$$ We have that $\tilde{N}\subseteq N$ and $\tilde{N}\unlhd_oG\rtimes H$: normality comes from the normality of $N$ in $G$ and from how we have defined $\tilde{N}$; it's open because we are doing a finite intersection of open subgroups (since they are conjugate to $N\leq_o G\rtimes H$). Now, suppose that $$G=\underset{i\in\mathbb{N}}{\varprojlim}\hspace{1mm}(\cdots\leftarrow G_i\leftarrow G_{i+1}\leftarrow\cdots)$$ with surjective connecting homomorphisms $\phi_i\colon G_{i+1}\to G_i$. So there is a descending chain of open normal subgroup $\{N_i\}_{i\in\mathbb{N}}$ (that is $N_i\unlhd_o G$, $N_{i+1}\subseteq N_i$ and $\bigcap_i N_i=1$) such that, for each $i\in\mathbb{N}$, $G/N_i\simeq G_i$ and the projection $\pi_i\colon G/N_{i+1}\to G/N_i$ induces $\phi_i$. Now, consider $\{\tilde{N_i}\}_{i\in\mathbb{N}}$: it is a descending chain of open normal subgroup of $G\rtimes H$ such that every $\tilde{N}_i$ is contained in $G$. In particular, we have that $$G=\underset{i}{\varprojlim}\hspace{1mm}\frac{G}{\tilde{N}_i}$$ and $$G\rtimes H=\underset{i}{\varprojlim}\hspace{1mm}\frac{G\rtimes H}{\tilde{N}_i}=\underset{i}{\varprojlim}\hspace{1mm}(\frac{G}{\tilde{N}_i}\rtimes H)$$ since $\frac{G\rtimes H}{\tilde{N}_i}\simeq\frac{G}{\tilde{N}_i}\rtimes H$. For every $i$, the map $\psi_i\colon\frac{G}{\tilde{N}_{i+1}}\rtimes H\to \frac{G}{\tilde{N}_i}\rtimes H$ is induced by the projection $\pi_i\colon G/\tilde{N}_{i+1}\to G/\tilde{N}_i$.