Let $G$ be a transitive subgroup of $S_4$.
By the orbit-stabilizer theorem, $4\mid|G|$.
Hence the possible order of $G$ is $4,8,12,24$.
So it is possible to list all the possible structure of $G$ by considering all the subgroups of $S_4$ with those orders.
However in An Introduction of Theory of Groups by Rotman page $69$ exercise $3.51$, it is done by considering the index $[G:G\cap V]$ where $V$ is the Klein $4$-group.
(i) If $m=[G:G\cap V]$, then $m\mid 6$
(ii) If $m=6$, then $G\cong S_4$; if $m=3$, then $G=A_4$; if $m=1$, then $G=V$; if $m=2$, then either $G\cong \Bbb{Z}_4$ or $G\cong D_8$.
I don't have idea to solve part (i) yet.
For part (ii), if $m=6$ then $G=S_4$ or $A_4$ by order consideration. But $A_4\cap V=V$, then $m=3$; a contradiction. Hence $G=S_4$.
If $m=3$, then similarly $G=S_4$ or $A_4$. But $S_4\cap V=V$, then $m=6$; a contradiction. Hence $G=A_4$.
If $m=1$,then $G=G\cap V\leq V$, hence $|G|\mid 4$. Thus $|G|=4$ and $G=V$.
So I am stuck in proving for the case $m=2$. So far I can note that $G\cap V \lhd G$.