$H\le G$. Number of orbits of the induced $H$-action of a transitive $G$-action: why assuming $H\unlhd G$? This post claims:

$G$ is a transitive group action. Normal subgroup of transitive group $G$ has at most $|G:N|$ orbits, and if $|G:N|$ is finite, then the number of orbits of $N$ divides $|G:N|$.

My question deals with the part in italics, as I seemingly proved it without using the normality of $N$. So I wonder whether I inadvertently used it somewhere, or is my proof just wrong (or, thirdly, the result is more general and holds for every subgroup $H\le G$, whereas perhaps the normality is used to prove the second part of the claim). Hereafter what I did:
Let $H\le G$. Every $g\in G$ lays in some right coset of $H$ in $G$. So, denoted with $R\subseteq G$ a complete set of coset representatives, for $x\in X$ we get:
\begin{alignat}{1}
\operatorname{Orb}_G(x) &= \{g\cdot x, g\in G\} \\
&= \{(hg_i)\cdot x, g_i\in R \text{ and } h\in H\} \\
&= \{h\cdot(g_i\cdot x), g_i\in R \text{ and } h\in H\} \\
&= \{h\cdot y_i, y_i\in Y(x) \text{ and } h\in H\} \\
&= \bigcup_{y_i\in Y(x)}\{h\cdot y_i, h\in H\} \\
&= \bigcup_{y_i\in Y(x)}\operatorname{Orb}_H(y_i) \\
\end{alignat}
where $Y(x):=\{g_i\cdot x, g_i\in R\}$. Now, $|Y(x)|\le |R|=[G:H]$, so if the $G$-action is transitive, then the $H$-action has at most $[G:H]$ orbits.
 A: Indeed, if $G\curvearrowright\Omega$ transitively then the (general version of the) orbit-stabilizer theorem says $\Omega\cong G/S$ as a $G$-set, where $S$ is the stabilizer of any element in $\Omega$ (these stabilizers are conjugate so the corresponding coset spaces are equivalent). Which in turn means if $H\le G$ then the orbit space $H\backslash \Omega$ is in bijection with the double coset space $H\backslash G/S$, and there is in turn a surjection to the double coset space from the right coset space $H\backslash G$, which makes $[G:H]$ an upper bound. This is basically a different perspective on your own proof.
So you're right the first part of the theorem doesn't require the hypothesis that $H$ is normal. But the second part of the theorem does require this hypothesis. If $N$ is normal, then the quotient group $G/N$ acts on the orbit space $N\backslash\Omega$, transitively because $G\curvearrowright\Omega$ was transitive, which by orbit stabilizer implies the size of the orbit space divides the size of $G/N$. But if $H$ is not normal, the $H$-orbits can have different sizes, and there is no guaranteed well-defined projection from $G/H$ to $H\backslash\Omega$ with equal-size fibers. And indeed, we can cook up a counterexample by looking at small groups presumably.
