Suppose we are given an action $\rho \colon (A \rtimes G) \times X \to X$. We define an $A$-Action $\rho_A\colon A \times X \to X$ by
\[
\rho_A(a, x) := \rho\bigl((a,1), x\bigr)
\]
and a $G$-action by
\[
\rho_G(g,x) := \rho\bigl((1,g), x\bigr)
\]
(Note that as $(1,g)(1,h) = (1,gh)$ and $(a,1)(b,1) = (ab,1)$ by definition of the semidirect product, so $\rho_G$ and $\rho_A$ are actions).
We have for $a \in A$, $g \in G$:
\begin{align*}
\rho_G\bigl(g,\rho_A(a,x)\bigr)
&= \rho\Bigl((1,g), \rho\bigl((a,1), x\bigr)\Bigr)
\\&= \rho\bigl((1,g)(a,1), x\bigr)
\\&= \rho\bigl((\sideset{^g}{}a,g), x\bigr)\\
\\&= \rho\bigl((\sideset{^g}{}a,1)(1,g), x\bigr)\\
\\&= \rho_A\bigl(\sideset{^g}{}a,\rho_G(g, x)\bigr)
\end{align*}
as wished.
The other way, given compatible actions $\rho_A\colon A \times X \to X$ and $\rho_G\colon G \times X \to X$, define $\rho \colon (A \rtimes G) \times X \to X$ by
\[ \rho\bigl((a,g), x\bigr) = \rho_A\bigl(a, \rho_G(g,x)\bigr)
\]
We will show, that this is an action, so let $(a,g), (b,h) \in A \rtimes G$, we have using the compability
\begin{align*}
\rho\bigl((a,g), \rho((b,h), x)\bigr)
&= \rho_A\biggl(a,\rho_G\Bigl(g, \rho_A\bigl(b,\rho_G(h,x)\bigr)\Bigr)\biggr)\\
&= \rho_A\biggl(a, \rho_A\Bigl(\sideset{^g}{}b, \rho_G\bigl(g, \rho_G(h, x)\bigr)\Bigr)\biggr)\\
&= \rho_A\bigl(a\sideset{^g}{}b, \rho_G(gh, x)\bigr)\\
&= \rho\bigl((a\sideset{^g}{}b, gh), x\bigr)\\
&= \rho\bigl((a,g)(b,h), x\bigr)
\end{align*}
As $(a,g)(b,h) = (a\sideset{^g}{}b, gh)$ in $A \rtimes G$.