Let $G$ act nonregular and transitive on $\Omega$ such that each nontrivial element has at most two fixed points.
Let $\alpha, \beta\in \Omega$ be distinct and such that $U := G_{\alpha}\cap G_{\beta} \ne 1$. Suppose that $|G_{\alpha}|$ is odd and that $|\Omega|$ is even. Suppose further that $G$ has no subgroup of index at most $2$ that is a Frobenius group. Then there exists an involution $x \in N_G(U) \setminus U$ that interchanges $\alpha$ and $\beta$.
Now suppose we have such an involution $x \in N_G(U) \setminus U$. Assume $C_U(x) \ne 1$ and set $C := C_G(x)$.
If $|\alpha^C| \le 2$, then $C = \langle x \rangle \times C_U(x)$ and $\langle x \rangle$ is a Sylow $2$-subgroup of $G$.
Why is $\langle x \rangle$ already a Sylow $2$-subgroup of $G$?
Also as $x$ interchanges $\alpha$ and $\beta$ we always have $x \notin C_U(x)$ and $x$ commutes with all elements from $C_U(x)$ by definition, so that $C = \langle x \rangle \times C_U(x)$ is always true, and not because $|\alpha^C| \le 2$. By other facts I see that $(|G_{\alpha}|, |\Omega|) = 1$, but not that $|\Omega|$ must have twice odd order.