I am fairly new to Group Theory. I know that $D_4$ is a combination of $C_4$ and $C_2$. Now, how to derive semidirect product for D4 group?
For instance, in $P_4$ group, if $t, p$ are translations, and $r, s$ are rotations, then $(t, r) \cdot (p, s) = (t + r \cdot p, r \cdot s)$.
Similarly, if I have an operation (f1, r1), where f1 is reflection and r1 is rotation, and another such operation (f2, r2), then how to figure out $(f_1, r_1) \cdot (f_2, r_2)$?
How can I come up with an equation for $D_4$, similar to how we have it for $P_4$?
As Arturo pointed out in the comments, in this setting the factor $C_2$ consists just of a single reflection $s$ together with the identity.
Here's a standard top-down approach: In general a semidirect product $N \rtimes H$ is specified by a homomorphism $\phi: H \mapsto \operatorname{Aut}(N)$, which determines the semidirect product multiplication $$(n_1, h_1) \cdot (n_2, h_2) := (n_1 \phi(h_1)(n_2), h_1 h_2) .$$ In our case, $N = C_4$, $H = C_2$---if we pick a generator $r$ of $C_4$, we have $N = \{e_N, r, r^2, r^3\}$ and $H = \{e_H, s\}$---so a semidirect product of the two groups is determined by a map $\phi: C_2 \mapsto \operatorname{Aut}(C_4) \cong C_4^{\times}$. There are two automorphisms of $C_4$: The identity automorphism $\iota$ and the "inversion" automorphism $\alpha : r^k \mapsto r^{-k}$ (for all $k$), which has order $2$.
Any homomorphism $\phi$ maps the identity $e_H$ to the identity automorphism $\iota$, and so $$(r^{k_1}, e_H) \cdot (r^{k_2}, s^\ell) = (r^{k_1} r^{k_2}, e_H s^\ell) = (r^{k_1 + k_2}, s^\ell) .$$
Thus, for $H = C_2$, $\phi$ is determined by $\phi(s)$. If $\phi(s) = \iota$, then $\phi$ is the trivial homomorphism, which yields the direct product. If instead $\phi(s) = \alpha$, the multiplication obeys $$(r^{k_1}, s) \cdot (r^{k_2}, s^\ell) = (r^{k_1} \phi(s)(r^{k_2}), s s^\ell) = (r^{k_1} r^{-k_2}, s^{\ell + 1}) = (r^{k_1 - k_2}, s^{\ell + 1}).$$ We can combine the previous two formulae into a general rule: $$\boxed{(r^{k_1}, s^{\ell_1}) \cdot (r^{k_2}, s^{\ell_2}) = (r^{k_1 + (-1)^{\ell_1} k_2}, s^{\ell_1 + \ell_2})} .$$ Of course, we can reduce the exponents of $r$ and $s$ modulo the orders of those elements, namely, $4$ and $2$ respectively.
If we write $(r^k, s^\ell)$ as $r^k s^\ell$, the multiplication is characterized by $s r = r^{-1} s$, which recovers the multiplication law and usual notation of the dihedral group $D_4$ (sometimes called $D_8$).