We have the following short exact sequence from the long exact sequence for a pair $$0\to\pi_2(\mathbb{S^2})=\mathbb{Z}\to\pi_2(\mathbb{S}^2,X)\to\pi_1(X)=F_2\to0.$$ I wanted to construct a section (I guess there is one), so $\pi_2(\mathbb{S}^2)=\mathbb{Z}\rtimes F_2$. What I did is that I tried to use the homotopy extension property for $(\mathbb{D}^2,\mathbb{S}^1)$ to say that I can map a homotopy class in $\pi_1(X)$ to maps that are still homotopic in the set of maps $\mathbb{D}^2\to\mathbb{S}^2$ and whose restrictions to $\mathbb{S}^1$ are also homotopic, so I get a section.
But say, the section $s:\pi_1(X)\to\pi_2(\mathbb{S}^2,X)$ maps a class $[f]\in\pi_1(X)$ to $[F]\in\pi_2(\mathbb{S}^2,X)$ where $F$ is an extension of $f$. Later I think not every map in $[F]$ restricted to $\mathbb{S}^1$ is homotopic to $f$... And if the homotopy extension property really works, the boundary map would always have a section, which seems absurd, though I am totally new to the relative homotopy groups... Even if the group is really a semidirect product, I have no idea how to figure out the action of $F_2$ on $\mathbb{Z}$. My guess is something like $\varphi\in\text{Aut}(\mathbb{Z})$, $\varphi(*^{d_1}*^{d_2}\cdots*^{d_k})=\sum d_i$, where $*^{d_1}*^{d_2}\cdots*^{d_k}$ denotes a word in $F_2$.
Hope to get some help... Thanks!
Since $\pi_2(A)$ is central in $\pi_2(B,A)$, it is never the nontrivial semidirect product. [See Spanier, Algebraic Topology, Chapter 7, Section 3, Corollary 13, page 386]
The fact that your sequence splits follows formally because $F_2$ is a free group.
Since you mentioned the more general question of the sequence splitting, if we have an exact sequence
$$0 \to \pi_2(A) \to \pi_2(A,X) \to \pi_1(X) \to 0$$
I think we can say the sequence is split if $X \to A$ is nulhomotopic ("$X$ is nulhomotopic in $A$"). This way you can choose a nulhomotopy and use it to make a filling-in of each path, and also a homotopy of their homotopies.
I think this agrees with Tyler Lawson's comment here https://mathoverflow.net/questions/294006/conditions-for-the-second-homotopy-group-to-be-abelian