Let $G$ be a finite, transitive, faithful group acting on $\Omega$ such that each nontrivial element has at most two fixed points.
Lemma: If $\alpha \in \Omega$ and $1 \ne X \le G_{\alpha}$, then for the subgroup $N_{G_{\alpha}}(X) = N_G(X) \cap G_{\alpha}$ we have $|N_G(X) : N_{G_{\alpha}}(X)| \le 2$.
Proof: Let $1 \ne X \le G_{\alpha}$ and $g \in N_G(X)$, then $X = X^g \le G_{\alpha^g}$, and so $\alpha^g$ is fixed by $X$, i.e. $N_G(X)$ acts on the fixed points of $X$. But $X$ has at most two fixed points, so either $N_G(X) \le G_{\alpha}$, or it induces the symmetric group on two points on the two fixed points of $X$, i.e. we have a surjective homomorphism $\rho : N_G(X) \to S_{\{\alpha,\beta\}}$, where $\beta$ denotes the other fixed point, with kernel $N_{G_{\alpha}}(X)$ and hence $|N_G(X) : N_{G_{\alpha}}(X)| = |S_{\{\alpha,\beta\}}| = 2$. $\square$
Now suppose we have $\alpha, \beta \in \Omega$ distinct, and $U := G_{\alpha} \cap G_{\beta} \ne 1$. Further let $|G_{\alpha}|$ be odd and $|\Omega|$ be even with $(|G_{\alpha}|,|\Omega|) = 1$ and suppose $G$ has no subgroup of index at most $2$ that is a Frobenius group. Then why this implies that there exists an involution $x \in N_G(U) \setminus U$?
The above Lemma and our hypothesis about $G$ that it does not contain a Frobenius group of index at most $2$ should somehow give that $|N_G(U) : U| = 2$, and this in turn should imply that there exists such an involution. But I do not see how these facts are connected?