(Ref: this) Let $\pi \times_{\varphi} G$ be semi-direct product in which $G$ is normal and $\pi$ is complement.
Let $\omega$ be another complement of $G$ in above semi-direct product (so $\pi \times_{\varphi} G= \omega \times_{\varphi} G $).
Givem $\sigma\in \pi$. This $\sigma$ may not be in $\omega$, but by multiplying by a suitable element $g\in G$ we can bring it in $\omega$, i.e. there is $g\in G$ such that $\sigma g\in\omega$. This $g$ is unique: for if $\sigma g\in \omega$ and $\sigma g_1\in \omega$ then $(\sigma g)(\sigma g_1)^{-1}\in \omega$, i.e. $\sigma (gg_1^{-1})\sigma^{-1} \in \omega$. Now $G$ is normal and $G\cap \omega=1$ implies that $g=g_1$.
Thus, we get a map $f\colon \pi\rightarrow G$, $f(\sigma)=g$. [it is called crossed homomorphism because-]
Fact: $f(\sigma\tau)=f(\sigma)^{\tau}f(\tau)$ [where symbol $x^y$ means $y^{-1}xy$].
I tried to look the map $f$ in a different way and this confused me too much:
Given $\sigma\in\pi$. We have obtained unique $g\in G$ such that $\sigma g\in \omega$. Then notice that $$\sigma = (\sigma g) g^{-1} =xg^{-1} \in \omega \times_{\varphi} G.$$ Call $\sigma g=x$ the $\omega$-component of $\sigma\in \omega \times_{\varphi} G$, and $g^{-1}$ the $G$-component.
From this, I understood $f$ as follows:
given $\sigma\in \pi$, write it as product $\omega$-component and $G$-component; this decomposition is unique in $\omega\times_{\varphi} G$ and then define $f(\sigma)$ to be inverse of the $G$-component of $\sigma$ in $\omega\times_{\varphi} G$.
This understanding of $f$ is confusing me to prove the Fact above. Because, if $\sigma,\tau\in \pi$, let $\sigma=xg^{-1}$ and $\tau=yh^{-1}$ (where $x,y\in\omega$ and $g,h\in G$). Then $f(\sigma)=g$ and $f(\tau)=h$. $$\sigma\tau = (xg^{-1})(yh^{-1}) = xy.y^{-1}g^{-1}y.h^{-1}=(xy).(hg^y)^{-1}.$$ From my understanding of $f$, I get $f(\sigma\tau)$ equal to inverse of $G$-component in last expression: $$f(\sigma\tau)=hg^y=f(\tau)f(\sigma)^{y}.$$ Question: Why we are getting a strange expression for $f(\sigma\tau)$ in last equation? I mean, in the Fact, the equation is completely in terms of $\sigma$ and $\tau$, but here the last expression, one strange factor appears is $y$. Further, last equation is not matching with that appearing in the Fact.
Thank you very much for the patience!
The expression for $f(\sigma\tau)$ you have derived is consistent with the one in the Fact.
From the definition of $y$ and $h$ we have $y = \tau h = \tau f(\tau)$, so you could write your expression as $f(\sigma\tau)=f(\tau)f(\sigma)^{\tau f(\tau)}$ to avoid mentioning $y$. Also
$\begin{align} f(\tau)f(\sigma)^{\tau f(\tau)} & = f(\tau)(\tau f(\tau))^{-1}f(\sigma)\tau f(\tau)\\ & = f(\tau)f(\tau)^{-1}\tau^{-1}f(\sigma)\tau f(\tau)\\ & = \tau^{-1}f(\sigma)\tau f(\tau)\\ & = f(\sigma)^{\tau} f(\tau). \end{align}$