If $H$ fixes three points, then could the normalizer of $H$ induce an orbit of size two on the fixed points Let $G$ be a transitive permutation group of degree $\ge 5$ acting such that every four-point stabilizer is trivial. Equivalently this means that every nontrivial element has at most three fixed points. Now if $1 \ne H \le G_{\alpha}\cap G_{\beta} \cap G_{\gamma}$, i.e. $H$ is a subgroup which fixes three points, this gives that
$$
 |N_G(H) : N_G(H) \cap G_{\alpha}| \le 3
$$
as $N_G(H)$ acts on the three fixed points.

I want to know if the case $|N_G(H) : N_G(H) \cap G_{\alpha}| = 2$
  is possible.

If $Z(G) \ne 1$, as $Z(G)$ similarly permutes the three fixed points, but could not fix any element itself (or otherwise if would fix everything, which is excluded as the action is faithful) it must move every fixed point and as $Z(G) \le N_G(H)$ we have $|N_G(H) : N_G(H) \cap G_{\alpha}| = 3$. But what if $Z(G) = 1$. Is it possible that $|N_G(H) : N_G(H) \cap G_{\alpha}| = 2$, which means that $\Delta = \{\alpha,\beta,\gamma\}$ is decomposed into two $N_G(H)$-orbits, one of size $2$ and one of size $1$.
Another condition where this is not possible is if the point stabilizers are odd, and if $|N_G(H) : N_G(H) \cap G_{\alpha}| = 2$
and $|N_G(H) : N_G(H) \cap G_{\gamma}| = 1$ would imply then $N_G(H) \cap G_{\gamma} = N_G(H)$, or $N_G(H) \le G_{\alpha}$ and so $N_G(H)$ would have odd order, contradicting $|N_G(H) : N_G(H) \cap G_{\alpha}| = 2$
Another condition where this is not possible is if distinct conjugates of the stabilizers intersect trivially, as then $H = G_{\alpha} = G_{\beta} = G_{\gamma}$ and so if $N_G(H) > H$ every element in the normalizer not in $H$ moves every point.
Okay, but what about the general case?

Yes this is possible. An example is ${\rm PSL}(3,2)$ in its natural action on $7$ points, with point stabilizer isomorphic to $S_4$, and $H$ a subgroup of order $2$ that does not lie in the derived subgroup of $G_\alpha$. Then $|N_G(H)|=8$, $|N_{G_\alpha}(H)|=4$.
