Let $G$ be a finite permutation group acting nonregular and transitive such that each nontrivial element fixes at most two points.
Lemma: (1) If $p$ is odd and divides the order of $G_{\alpha}$, then $G_{\alpha}$ contains a Sylow $p$-subgroup of $G$;
(2) If $1 \ne X \le G_{\alpha}$, then $|N_G(X) \cap G_{\alpha} : N_G(X)| \le 2$
Proof: The first follows by considering the $P$-orbits on $\Omega$ and thereby showing that $|G : G_{\alpha}|$ is not divisible by $p$, the second follows by noting that $N_G(X)$ acts on the at most two fixed point of $X$ and has point stabilizer $N_{G_{\alpha}}(X) = N_G(X) \cap G_{\alpha}$. $\square$
Statement: Suppose that $N$ is a proper normal $2'$-subgroup of $G$, and that $\alpha \in \Omega$. Suppose that $p \in \pi(N)$ is such that $P := O_p(N) \ne 1$. Then every $x \in G_{\alpha}, x \ne 1$ acts fixed point freely on $P$. If we let $\overline \Omega$ denote the set of $P$-orbits on $\Omega$ and $\overline G := G/P$, then one of the following is true:
(1) $|\overline \Omega| = 1$ and $G = PG_{\alpha}$ is a Frobenius group.
(2) $|\overline \Omega| = 2$ and $PG_{\alpha}$ is a Frobenius group of index $2$ in $G$.
(3) We have $|\overline \Omega| \ge 4$ and $\overline G$ acts faithful, transitive, non-regular on $\overline \Omega$ such that each nontrivial element fixes at most two points.
Proof: If $p$ divides $|G_{\alpha}|$, then the above Lemma (1) yields that $P \le G_{\alpha}$ and hence $P$ fixes every point in $\Omega$, because $P \unlhd G$. This is impossible and therefore $p$ does not divide $|G_{\alpha}|$. If follows that $P$ has only regular orbits (i.e. acts semi-regularly) on $\Omega$. As $p$ is odd and $G_{\alpha}$ is a $p'$-group now, it follows by the above Lemma (2) that $G_{\alpha}$ acts fixed point freely on $P$. Now we consider the action of $\overline G$ on $\overline \Omega$. If $|\Omega| = 1$, then $\Omega = \alpha^P$ and $G = PG_{\alpha}$. As $G_{\alpha}$ acts fixed point freely on $P$, it follows that $G$ is a Frobenious group with complement $G_{\alpha}$, proving (1). If $|\overline \Omega| = 2$, then $PG_{\alpha}$ is of index $2$ in $G$, hence normal in $G$. The action of $G_{\overline \alpha}$ on $\alpha^P$ is as above and thus $G_{\overline \alpha}$ is a Frobenious group, and (2) follows. So we may now assume that $|\overline \Omega| > 2$. Let $\overline \omega_1, \overline \omega_2 \in \overline \Omega$ with representatives $\omega_1, \omega_2 \in \Omega$. Then there exists $g \in G$ such that $\omega_1^g = \omega_2$ and $\overline \omega_1^{\overline g} = \overline \omega_2$. As $G_{\alpha} \ne 1$, we may take $x \in G_{\alpha}, x\ne 1$ and we see that $x$ fixes $\overline \alpha$, so $\overline G$ does not act regularly on $\overline G$.
Let $\overline g \in \overline G\overline , g \ne \overline 1$ and suppose that $\overline w$ is a fixed point of $\overline g$ in $\overline \Omega$. As $\overline G_{\overline \omega} = G_{\omega}P/P$ we may suppose that $g \in G_{\omega}$. As $P$ acts regularly on every element of $\overline \Omega$, it follows that $g$ acts on $\overline \omega$ in the same way as on $P$, hence with a unique fixed point. This means that $g$ has a fixed point in every $P$-orbit that it stabilizes. Our hypothesis yields that $g$ stabilizes at most two $P$-orbits and hence $\overline g$ fixes at most two element of $\overline \Omega$. $\square$
1) As $p$ is odd and $G_{\alpha}$ is a $p'$-group now, it follows by the above Lemma (2) that $G_{\alpha}$ acts fixed point freely on $P$
I do not see that. We have to show that for each nontrivial $x \in G_{\alpha}$ we have $u^x \ne u$ for each $u \in P$ by Lemma (1). The only way I can think of is to consider $X := \{ x \in G_{\alpha} : u^x = u \}$ for $u \in P$, i.e. a point stabilzer in the action of $G_{\alpha}$ on $P$. Suppose $X \ne 1$, then Lemma (1) gives $$ |N_G(X) \cap G_{\alpha} : N_G(X)| \le 2 $$ and of course $|G_{\alpha}| = |N_G(X) \cap G_{\alpha} : N_G(X)| \cdot ||N_G(X) \cap G_{\alpha}|$. But I do not see where this might be a contradiction if $G_{\alpha}$ is a $p'$-group if $p$ is odd?
2) If $|\Omega| = 1$, then $\Omega = \alpha^P$ and $G = PG_{\alpha}$.
Why does $G = PG_{\alpha}$ follows? Guess we have to show that $G / P = G_{\alpha}P/P$, but do not know how?
3) If $|\overline \Omega| = 2$, then $PG_{\alpha} = G_{\overline \alpha}$ is of index $2$ in $G$.
Why has $PG_{\alpha} = G_{\overline \alpha}$ index $2$? By orbit-stabilizer we have $|\overline G : \overline G_{\overline \alpha}| = 2$ and $|\overline G : \overline G_{\overline \alpha}| = |G/P : G_{\alpha}P/P|$. But this equals $|G : G_{\alpha}P|$ just in case $G_{\alpha}P$ is normal by the second isomorphism theorem, and we do not know that it is normal yet (before knowing that its index is two).
I hope someone can explain this to me?! Point 1) I find most obscure.