Is the commutator subgroup of a subgroup the same as the commutator subgroup of the group intersected with that subgroup?

This is false. Take $G$ a non-abelian simple group. Then its commutator subgroup is all of $G$. So letting $H$ be any abelian (edit: oops! thanks to the commentors) subgroup of $G$ gives a counterexample.

Another small counterexample: Let $G$ be the group of permutations of $3$ objects (the smallest nonabelian group) and let $H$ be its cyclic subgroup of order $3$. The commutator subgroup of $G$ is $H$ but the commutator subgroup of $H$ is trivial.