If $U,V \subset S_n$ are subgroups with $S_n//U = \{id,g_2,...,g_e\}$ and $\alpha_j$ is $\frac{1}{j}$ times the number of $i\in [e]$ s.t $[V:V \cap g_i U g_i^{-1}]=j$ then $(\alpha_1,...,\alpha_e)$ is a partition of $e$.
Now let $U=V \in \{V_4,D_4\}$ and $n=4$ ($V_4$ is the group with only the double transpositions). Then the number of $V_4$-orbits of $S_4/V_4$ of size $1$ is $6$, and there are none of greater size. One $D_4$-orbit of $S_4/D_4 = \{D_4,(12)D_4,(14)D_4 \}$ is $\{D_4\}$; for the other two, we have $|orb_{D_4}((ab)D_4)|=[D_4 : D_4 \cap (ab)D_4(ab)]=[D_4:V]=2$, so there is one orbit of size $1$ and two of size $2$.
I don't understand what I've done wrong here. I got a partition of $[S_4 :V_4]$, $1^6$, but apparently something went wrong with $D_4$, but I have no idea why? In the article which all this stuff pertains to it is asserted that there is one orbit of size one and one of size two, when $D_4$ acts on $S_4/D_4$. It cannot be that the two orbits should be considered as one, for if so then there would only be one orbit of size one wrt $S_4/V_4$.
(This is one of those questions that's difficult to pose because there's too much background, but I think this info should do, hopefully. Otherwise, if you're feeling more helpful than usual, check out beginning of section 3 and Table 2 here.)
Edit: Some calculations. $orb_V(V)={V}$, trivial ($V=V_4$). $orb_V((12)V)=\{(12)V,(34)V,(1324)V,(1423)V \}$ and $orb_V((13)V)=\{(13)V, (1234)V, (24)V, (1432)V\}$. To see that each of these orbits is really just $\{V\}$, use $\sigma V = \tau V \iff \tau^{-1} \sigma \in V$. Thus we have three orbits of size one so far.
With $D_4$, we have $orb_{D_4}((12)D_4)=\{ (12)D_4, (134)D_4, (243)D_4, (123)D_4, (142)D_4, (34)D_4, (1423)D_4, (1324)D_4\}$ and $orb_{D_4}((14)D_4) = \{ (14)D_4, (234)D_4, (132)D_4, (143)D_4, (124)D_4, (1342)D_4, (1243)D_4, (23)D_4 \}$. Some manipulations show that these two orbits have size two, yet only one of them counts?
There are a few unstated conventions. Which $D_4$? How do we multiply permutations? I'll go with GAP conventions for both as they are likely to agree with the paper and most of the world of computational group theory and finite group theory.
Take $U=D_4 = \{ (), (3,4), (1,2), (1,2)(3,4), (1,3)(2,4), (1,3,2,4), (1,4,2,3), (1,4)(2,3) \}$ and multiply permutations in reading order such that $(1,2)(2,3)=(1,3,2)$.
$G//U=S_4//D_4 = \{ ()D_4, (1,2,3)D_4, (1,3,2)D_4 \}$ so $e=3$.
If $V=D_4$ as well, then
$V \cap g_1Ug_1^{-1} = V \cap ()U()^{-1} = V \cap V = V$ has index 1. Note that the orbit of $D_4$ on the seed $()D_4$ is $\{ D_4 \}$ of size 1.
$V \cap g_2 U g_2^{-1}$ has size 4 (it is $V_4$, the Klein viergruppe of double transpositions). That means it has index 2. Note that the orbit of $D_4$ on the seed $(1,2,3)D_4$ is $\{ (1,2,3)D_4, (1,3,2)D_4 \}$.
$V \cap g_3 U g_3^{-1}$ has size 4 (it is $V_4$, the Klein viergruppe of double transpositions). That means it has index 2. Note that the orbit of $D_4$ on the seed $(1,3,2)D_4$ is $\{ (1,2,3)D_4, (1,3,2)D_4 \}$.
Now we get $\alpha_1 = \tfrac{1}{1} 1 = 1$ since only $g_1$ gave index 1.
We get $\alpha_2 = \tfrac{1}{2} 2 = 1$ since $g_2$ and $g_3$ gave index 2.
We get $\alpha_3=\tfrac{1}{3} 0 = 0$ since nothing gave index 3.
Sure enough $1 \alpha_1 + 2\alpha_2 + 3 \alpha_3 = 1 + 2 + 0 = 3$ is a partition.
In terms of orbits, this is because the three cosets $\{ ()D_4, (1,2,3)D_4, (1,3,2)D_4 \}$ split into two orbits under the left action of $D_4$, namely, $\{ ()D_4 \}$ and $\{ (1,2,3)D_4, (1,3,2)D_4 \}$.