"Let $N := O^{2'}(G)$, then $N \unlhd G$ is a normal subgroup, and suppose that $C_G(N) = G$. Then $G$ has a central Sylow $2$-subgroup. So $G$ has a normal $2$-complement."
I do not understand the above reasoning. Why is the Sylow $2$-subgroup central? And why does this implies that the $2$-complement is normal?
On
Since $C_G(N)=G$ , $N\leq Z(G)$ as it means that every element in $G$ commutes every element in $N$.
Now, By schurzasenaus theorem, $G=NH$ and any other complement is in the form $H^g$.
Let $g=nh$ then $H^g=H^{nh}=H^n=H$ the last equality is because of $N\leq Z(G)$.
Thus, $H$ is also normal in $G$, we have $G\cong H\times N$
Let $P \in Syl_2(G)$, then $PN/N \in Syl_2(G/N)$. But $G/N$ has odd order, so it follows that $PN=N$, that is, $P \subseteq N$. But $C_G(N)=G$ is equivalent to $N \subseteq Z(G)$, so $P \subseteq Z(G)$, as wanted.
$P$ is a normal subgroup and Sylow, so by Schur-Zassenhaus, $G=PH$, for some complement $H$ of $P$. Of course, $H$ normalizes itself, and $P$, being central, normalizes $H$. Hence $G=PH$ normalizes $H$, in other words, $H$ is a normal subgroup and it follows that $G \cong P \times H$. Finally, since $|H|$ is odd, $P=N$.