Expanding PavelC's comment into an answer:
$S_4$ has 2 normal subgroups: $A_4$, and $V$, the group of double transpositions. $V\subset A_4$, so if you take $G$ to be $S_4$, your $H$ is $V$. $V$ is isomorphic to the Klein 4-group, so it is not simple.
What's going on in this case (the substance of your observation that normality is not transitive) is that subgroups of $V$ are not normal in $S_4$ even though they are normal in $V$. More specifically, the subgroups of $V$ cannot be conjugated to each other inside $V$ since it is abelian; therefore they are all normal in $V$. However, they are all conjugate inside $G$. For example, one of the subgroups of $V$ is the order 2 cyclic group generated by $(12)(34)$ (call it $J$). This subgroup is normal in $V$ since $V$ is abelian, but we have
$$(13)(12)(34)(13) = (14)(23)$$
and therefore $(13)J(13)$ is the subgroup generated by $(14)(23)$. Similarly,
$$(23)(12)(34)(23) = (13)(24)$$
and therefore $(23)J(23)$ is the subgroup generated by $(13)(24)$. Thus all three cyclic subgroups of $V$ are conjugate in $S_4$.