I think I can start from Normal subgroups of D6: $<r>$,$<s,r^2>$, $<sr, r^2>$, and $<s,r^3>$. Because the other normal subgroups $<r^2>$ and $<r^3>$ do not make Simple groups when I do D6/$<r^2>$ or D6/$<r^3>$ (How do I show this?).
And so starting from D6, $<r>$, I can go to $<r^2>$ or $<r^3>$.
Starting from D6, $<s,r^2>$, I can only go to $<r^2>$ because it is only normal subgroup of $<s,r^2>$.
Starting from D6, $<sr,r^2>$, I can only go to $<r^2>$ because it is only normal subgroup of $<s,r^2>$.
Starting from D6, $<s,r^3>$, I can to $<s>$ or $<r^3>$ because both are normal, and If I divide, I get simple groups cycle 2 and 2.
Is this OK? How do I know this is all? Thank you! I saw this Composition Series for Dihedral Groups but I didn't learn semi-direct products so didn't understand.
First off, $\langle s, r^3 \rangle$ is not a proper normal subgroup, as it would also have to contain $rsr^{-1} = r^2s$ and then it would also have to contain $r$ itself and it would contain the entire group.
Now, as for your first question, note that $D_6/\langle r^2\rangle \cong D_3$ and that $D_6/\langle r^3 \rangle \cong D_2$ and that each of these have normal subgroups $\mathbb{Z}_3$ and $\mathbb{Z}_2$ respectively.
And as for how you know this is all, well, if you think about your algorithm, it was to look at every single possible normal subgroup of a group, determine if the composition factor is simple, and if so, continue to that subgroup. By definition of composition series, this will generate all of them.