Consider $F_2=\langle a,b \rangle$, the free group on $2$ letters.
If I look at $H=\langle ab,ba^{-1}, bab^{-1} \rangle$ (the generators are arbitrary), since each generator can be made from $a$ and $b$, $H \subseteq F_2$ and by definition of the presentation, it's a group so $H \le F_2$ ($H$ is a subgroup of $F_2$).
When exactly can I not use this logic? Specifically, is it true if I simply have that "the generating set of $H$ can be made by the generators of $G$", that $H \le G$?
I don't think it is true, because we could look at $H = \langle a, a^4=1\rangle$, and even though the generating set of $H$ sits inside that of $F_2$, $H$ is a finite group and not a subgroup of $F_2$ (finite groups are non-free and subgroups of free groups are free).
Then I thought perhaps the generators and the relators in $H$ must be derivable from those in $G$, which in the case of $F_2$ would mean that any subgroup should have generators made by $a,b$, and no nontrivial relators (i.e. the relators in the subgroup must be the trivial relation in the group) , but that is kind of assuming that the subgroups of free groups are free.
So in essence, given $H= \langle S_H \ | \ R_H \rangle$ and $G= \langle S_G \ | \ R_G \rangle$, when can we say $H \le G$?