Doubt on why in $D_{n}\;H\nsubseteq R=\langle r\rangle$ implies $\langle r^{k}\rangle;\langle sr^{h}\rangle$ subsets of $H$

Question

I have an exercise which consists on "meditating about the subgroups of $D_{n}$", so I considered the theory which I think I understood except one point. To study the subgroups of $D_{n}$ we shall consider $2$ cases given $H\le D_{n}$: $H\subseteq R$,$H\nsubseteq R$. The first is trivial as $R=\langle r\rangle,r=\frac{2\pi}{n}$ rotation, is cyclic and so is $H\subseteq R$; by Lagrange's theorem $|H|\;|n=|R|$ and $H$ is the only subgroup of $R$ with this property; $H=\langle r^{\frac{n}{d}}\rangle$ where $d|n$ and we have described $H\subseteq R$. The second is harder, but $[D_{n}:R]=2\rightarrow R\trianglelefteq D_{n}$ and so it makes sense to consider $D_{n}/R$ a group, we also know by that criterion $D_{n}/R\simeq \mathbb{Z}_{2}$. As we have a quotient group let's consider the canonical projection $\pi_{R}:D_{n}\rightarrow D_{n}/R:g\mapsto gR$; $H\nsubseteq R\rightarrow \exists h\in H|h\notin R\rightarrow \pi_{R}(h)\ne R\rightarrow \pi_{R}(h)\nsubseteq R$. But as we said $D_{n}/R\simeq \mathbb{Z}_{2}$ so $D_{n}/R$ has $2$ subgroups, the trivial ones: $D_{n}/R;\{R\}$. It follows $\pi_{R}(H)=D_{n}/R$. We continue by studying $ker(\pi_{R|H})=ker(\pi_{R})\cap H:=R\cap H$. Using the first homomorphism theorem for groups: $\frac{H}{H\cap R}\simeq \mathbb{Z}_{2}\rightarrow |H\cap R|=\frac{1}{2}|H|$. $\;H\cap R\subseteq R\rightarrow \exists k\in \mathbb{Z}|\;H\cap R=\langle r^{k}\rangle$. This step is clear as we simply used that $R\cap H$ is inside a cyclic subgroup; but now, the conclusion is: $\langle r^{k}\rangle$; $\langle sr^{h}\rangle[h\in \mathbb{Z}]$ are subsets of $H$. I'm wondering where $\langle sr^{h}\rangle$ came from. Is it simply because $\langle sr^{h}\rangle$ is "outside" $R$ and we asumed $H\nsubseteq R$? That seems intuitive, but why $\langle sr^{h}\rangle$ should be really a subset of $H$ is not clear to me.