We looked at this question during our research retreat and obtained the following characterisation: If $H$ is transitive on $X$ then it will be generated by its point stabilisers if and only if it does not have a proper system of imprimitivity upon which it acts regularly. By proper I mean one with at least two parts and by regular I mean that any element of $H$ that fixes one block in the partition fixes all of the blocks in the partition.

The proof is as follows: Let $N$ be the normal subgroup of $H$ that is generated by all the point stabilisers. If $N\neq H$ then since $G_x\leqslant N< H$ for any $x\in X$ it follows that $N$ is intransitive and so its set of orbits forms a proper system of imprimitivity. If $B$ is one orbit of $N$ then $H_B$ acts transitively on $B$. Since $N$ also acts transitively on $B$ we have that $H_B=NH_x$ for some $x\in B$. Thus $H_B=N$ and so $H_B$ fixes each $N$-orbit, that is, $H$ acts regularly on the set of $N$-orbits. Conversely, suppose that $\mathcal{B}$ is a proper system of imprimitivity upon which $H$ acts regularly. Let $K\neq H$ be the kernel of this action. Let $x\in X$ and $B$ be the block of $\mathcal{B}$ containing $x$. Then $H_x\leqslant H_B=K$. Since $K$ is normal and all point stabilisers are conjugate to $H_x$ it follows that $K$ contains all point stabilisers and so the subgroup generated by all point stabilisers is contained in $K$ and so is not equal to $H$.

We looked at this question during our research retreat and obtained the following characterisation: If $H$ is transitive on $X$ then it will be generated by its point stabilisers if and only if it does not have a proper system of imprimitivity upon which it acts regularly. By proper I mean one with at least two parts and by regular I mean that any element of $H$ that fixes one block in the partition fixes all of the blocks in the partition.

The proof is as follows: Let $N$ be the normal subgroup of $H$ that is generated by all the point stabilisers. If $N\neq H$ then since $H_x\leqslant N< H$ for any $x\in X$ it follows that $N$ is intransitive and so its set of orbits forms a proper system of imprimitivity. If $B$ is one orbit of $N$ then $H_B$ acts transitively on $B$. Since $N$ also acts transitively on $B$ we have that $H_B=NH_x$ for some $x\in B$. Thus $H_B=N$ and so $H_B$ fixes each $N$-orbit, that is, $H$ acts regularly on the set of $N$-orbits. Conversely, suppose that $\mathcal{B}$ is a proper system of imprimitivity upon which $H$ acts regularly. Let $K\neq H$ be the kernel of this action. Let $x\in X$ and $B$ be the block of $\mathcal{B}$ containing $x$. Then $H_x\leqslant H_B=K$. Since $K$ is normal and all point stabilisers are conjugate to $H_x$ it follows that $K$ contains all point stabilisers and so the subgroup generated by all point stabilisers is contained in $K$ and so is not equal to $H$.