Let $G$ be a group and $H$, $K$ ($K$ is normal) two subgroups of $G$ but neither $H$ is subgroup of $K$ nor $K$ is subgroup of $H$. What would be the problem with $H/K$? If I define for $x,y \in H$ multiplication as $xKyK=xyK$ then it is well defined?
Suppose $aK=xK$ and $bK=yK$ so $a^{-1}x \in K$ and $b^{-1}y \in K$. Then $b^{-1}a^{-1}xy=b^{-1}yy^{-1}a^{-1}xy \in K$. So the multiplication in $H/K$ is well defined no?
On
You can think of this in terms of homomorphisms. That is, by the first isomorphism theorem there exists a homomorphism $\phi: G\rightarrow \overline{G}$ such that $\ker\phi=K$ (and, of course, $\overline{G}$ is really just $G/K$). Then what you are after is $\phi(H)$. (This has the benefit of "making sense", while $H/K$ is a slightly dubious concept as for the notation to make sense you really need $K\leq H$.)
All you are really doing is proving that multiplication in $\phi(H)$ is well-defined. Which it is, as $\phi$ is a homomorphism.
This is fine. What you are really defining is multiplication in $HK/K$, which is a subgroup of $G/K$. By the second isomorphism theorem, this is isomorphic to $H/(H\cap K) $.