Hints:
Prove that
$$R\subset G\times G\;,\;\;(x,y)\in R\iff \;\exists \;h\in H\;,\;h'\in H'\;\;s.t.\;\;x=hyh'$$
is an equivalence relation on $\,G\;$ , and thus $\;G\;$ is the disjoint union of this equivalence relation's equivalence classes.
Now, what are the equivalence classes? :)
$1$.) I count 16 as well:
$D_{12}$ itself and the identity (trivial) subgroup.
$\langle r\rangle,\langle r^2\rangle, \langle r^3\rangle$
Two isomorphs of $S_3$: $\langle r^2,s\rangle$ and $\langle r^2,rs\rangle$.
Three isomorphs of $V$: $\langle r^3,s\rangle$, $\langle r^3,rs\rangle$ and $\langle r^3,r^2s\rangle$.
Six subgroups generated by a single reflection $\langle r^ks\rangle$ for $k = 0,1,2,3,4,5$.
$2$.) Yes, you want to find the normalizers in $D_{12}$. This is the entire group (as these are all normal) for the first seven subgroups listed, which greatly simplifies things. The isomorphs of $V$ are non-normal but of prime index, so they must be their own normalizers. It is clear that the normalizers of the reflection-generated subgroups of order 2 must be an isomorph of $V$ (the isomorphs of $V$ are abelian so they normalize any subgroup), as $S_3$ has no normal subgroups of order 2 (we are using the fact that the normalizer of a subgroup $H$ in a group $G$ is the largest subgroup of $G$ containing $H,$ in which $H$ is normal).
$3$.) Finding the orbits by brute force is not so bad: the seven normal subgroups will all have a single element of just themselves in their orbits, because they are stabilized by all of $D_{12}$. From Sylow theory (or by inspection), we see the isomorphs of $V$ are all conjugate, so there is another orbit (remember conjugation preserves order of subgroups), and we are left with finding the orbits of the reflection-generated subgroups.
By the orbit-stabilizer theorem (or by examining their generators' (ordinary) conjugacy classes) we have that these must occur in two orbits of three: $\{\langle s\rangle,\langle r^2s\rangle,\langle r^4s\rangle\}$ and $\{\langle rs\rangle,\langle r^3s\rangle,\langle r^5s\rangle\}$
Thus: $G/X = \{\{\{1\}\},\{\langle r\rangle\},\{\langle r^2\rangle\}, \{\langle r^3\rangle\},\{D_{12}\},\{\langle r^2,s\rangle\},\{\langle r^2,rs\rangle\},\{\langle r^3,s\rangle, \langle r^3,rs\rangle,\langle r^3,r^2s\rangle\},\{\langle s\rangle,\langle r^2s\rangle,\langle r^4s\rangle\},\{\langle rs\rangle,\langle r^3s\rangle,\langle r^5s\rangle\}\}.$
(This is a set with 10 elements).
Best Answer
We have to count the number of left cosets $xH$ in $G/H$ such that $xH=HxH$. Let $$H'=\{x\in G:xH=HxH\}.$$ It is easy to check that $H'$ is a group containing the group $H$. It suffices to show that $H'\ne H$ because then $|H':H|\ge p$, so there are at least $p$ distinct cosets $xH$ with $x\in H'$.
We prove the required claim by induction on $|G|$ as follows. Suppose that $|G|=p$. Since $H$ is a proper subgroup of $G$, $H=\{e\}$, so there are exactly $p$ one-element left cosets $xH$ and each of them is double.
Suppose that the claim is proved for all $p$-groups with smaller orders than $|G|$. Let $Z$ be the center of $H$. Clearly, $Z\subset H'$. So if $Z\not\subset H$ then we are done. Thus we suppose that $Z\subset H$. Then $H/Z$ is a proper subgroup of $G/Z$. It is well-known (see, for instance [Lan, I.6.$\S 3$]) that $Z\ne\{e\}$, so $|G/Z|<|G|$. By the inductive hypothesis, there exists a class $\bar x\in G/Z\setminus H/Z$ such that $\bar x (H/Z)=(H/Z)\bar x (H/Z)$. Let $x$ be any element of $\bar x Z$. Then $x\not\in H$, but $HxH\subset H\bar xH \subset \bar xH=xH,$ so $x\in H'$.
References
[Lan] Serge Lang, Algebra, Addison-Wesley, Reading, Mass., 1965 (Russian translation, Moskow, 1968).