Take $U$ a finite dimensional representation of $G\times H$. Consider only the $G$ action. Then every element from $H$ is an intertwining operator. Decompose $U$ into $G$-isotypic components. Then $H$ invariates each of them. Note that for every irreducible representation of $G$ the intertwining operators are scalers. Therefore, each $G$ isotypic of type $\rho$ of $U$ can be written as $\rho\otimes \eta$, where $\eta$ is a representation of $H$. Therefore, each finite dimensional (continuous) representation of $G\times H$ can be written as a direct sum of $\rho_1\otimes \rho_2$, where $\rho_1$ is an irrep of $G$ and $\rho_2$ is an irrep of $H$. From here it is easy also to show that $\rho_1\otimes \rho_2$ are irreducible.
This argument avoids most of the analysis. It only uses Schur's lemma on intertwining operators. One needs to argue that $\eta$ above is continuous, if we started with a continuous representation of $G\times H$, which is straightforward, looking at matrix coefficients.
In this answer I will try to explain in details the proof of
Proposition 25 in Serre's book. For my convenience in writing this,
I will use Serre's notation so my $A$ is your $N$, i.e.
$G=H\ltimes A$ and $G_i=H_i\ltimes A$ where $H_i=\{h\in H:h\chi_i=\chi_i\}$
subgroup of $H$.
(Serre) The restriction of $\theta_{i,\rho}$ to $A$ only
involves characters $\chi$ belonging to the orbit $H\chi_i$
of $\chi_i$. This shows that $\theta_{i,\rho}$ determines $i$.
As the statement said, we consider character $\chi$ of $\theta_{i,\rho}$
for $a\in A$. Recall the character for induced representation for $s\in G$,
$H$ subgroup of $G$, $f:H\to \mathbf{C}^{\times}$, is
$$\text{Ind}_H^G(f)(s)=f'(s)=\frac{1}{|H|}\sum_{t\in G, t^{-1}st\in H}f(t^{-1}st)$$
Hence,
\begin{align*}
\chi(a) & = \frac{1}{|G_i|}\sum_{t\in G, t^{-1}at\in G_i}
\chi_{\chi_i\otimes \tilde{\rho}}(t^{-1}at), \\
& = \frac{1}{|G_i|}\sum_{t\in G} \chi_i(t^{-1}at)\chi_{\tilde{\rho}}(t^{-1}at),
\; (A \text{ normal so } t^{-1}at\in A\subset G_i \; \forall t\in G), \\
& = \frac{\text{dim }V}{|G_i|}\sum_{t\in G}\chi_i(t^{-1}at), \;
(t^{-1}at\in A \Rightarrow \tilde{\rho}(t^{-1}at)=\rho(1)=\text{id}_V), \\
& = \frac{\text{dim }V}{|G_i|} \sum_{a'\in A,h\in H}\chi_i((a'h)^{-1}a(a'h)),\\
& = \frac{\text{dim }V}{|G_i|} \sum_{a'\in A,h\in H} \chi_i(h^{-1}ah), \;
(A \text{ abelian}) \\
& = \frac{|A|\text{dim }V}{|G_i|} \sum_{h\in H} (h\chi_i)(a).
\end{align*}
Note that $h\chi_i\in X=\text{Hom}(A,\mathbf{C}^{\times})$ is
an irreducible representation of $A$ so the above implies that
the restriction of $\theta_{i,\rho}$ to $A$ is the direct sum of
representations corresponding to $h\chi_i$. Note that $h\chi_i$'s lie
in the orbit $H\chi_i\subset X$, which is disjoint with $H\chi_{i'}$
for $i\ne i'$. Hence, if $i\ne i'$, the restriction of $\theta_{i,\rho}$
and $\theta_{i',\rho'}$ to $A$ are not isomorphic.
(Serre) Let $W$ be the representation space for $\theta_{i,\rho}$, let
$W_i$ be the subspace of $W$ corresponding to $\chi_i$ [the set
of $x\in W$ such that $\theta_{i,\rho}(a)x=\chi_i(a)x$ for all
$a\in A$]. The subspace $W_i$ is stable under $H_i$ and the representation
of $H_i$ in $W_i$ is isomorphic to $\rho$; whence $\theta_{i,\rho}$
determines $\rho$.
Recall from $\S 7.1$ of Serre's book, $W$ can be identified with
$W=\mathbf{C}[G]\otimes_{\mathbf{C}[G_i]}V$ where
$V$ is the representation space of $\chi_i\otimes \tilde{\rho}$.
In particular, this is a $\mathbf{C}[G]$-module where $g$ acts
on $g'\otimes v$ by $(gg')\otimes v$. Also note that,
$(gg_i)\otimes v$ and $g\otimes (\chi_i\otimes \tilde{\rho})(g_i)v$
are considered the same in $W$ for $g_i\in G_i$.
Now, we identify $W_i$. For $a\in A, g\otimes v\in W$, we have
$\theta_{i,\rho}(a)(g\otimes v)= (ag)\otimes v
= g \otimes (g^{-1}ag)v$. Since $g^{-1}ag\in A$ so
$(\chi_i\otimes \tilde{\rho})(g^{-1}ag)v=\chi_i(g^{-1}ag)v$.
Thus, $\theta_{i,\rho}(a)(g\otimes v)= \chi_i(g^{-1}ag)
(g\otimes v)$. Thus, in order for
$\theta_{i,\rho}(a)(g\otimes v)=\chi_i(a)(g\otimes v)$ for all
$a\in A$, we must have $\chi_i(a)=(g\chi_i)(a)$ for all $a\in A$.
This follows $g\in H_i$. Thus, $W_i=\mathbf{C}[H_i]
\otimes_{\mathbf{C}[G_i]} V$.
With this, it is obvious that $W_i$ is stable under $H_i$.
Furthermore, observe $W_i$ is spanned by $1\otimes v_j$
where $v_j$'s basis of $V$, $1$ identity in $H_i$ (this holds
since $h\otimes v=1\otimes \rho(h)v$ for all $h\in H_i,v\in V$).
Hence, one can easily construct isomorphism representation of
$H_i$ in $W$ to $\rho$, as desired.
Best Answer
I am not aware of a general procedure that would work for any semi-direct product. Serre treats semi-direct products by abelian groups in Part II, Section 8.2. See also my answer to another question. In your particular case, apart from lifting the representations of $H$ from the quotient, I would also try to induce the irreducible representations of $SL_2(\mathbb{F}_q)$ to $G$ and see which ones remain irreducible or where you can split off summands that you already know about. Mackey's irreducibility criterion (see e.g. Serre, Part II, section 7.4) should be quite useful for this. Generally, taking inner products of two such induced characters is easy using Frobenius reciprocity and Mackey's formula.