I am reading the proof here: https://proofwiki.org/wiki/Orbit-Stabilizer_Theorem
In the last lines, it says that $\text{Orb}(x)$ has the same number of elements as $G / \text{Stab} (x)$, So $|\text{Orb} (x)| = |G / \text{Stab} (x)|$. But it also says that $|G / \text{Stab}(x)| = [G: \text{Stab} (x)]$, which means that the order of the quotient group is equal to the index.
However, on the Wikipedia page for "Index of a subgroup" it says that if $N$ is a normal subgroup of $G$, then the index of $N$ in $G$ is also equal to the order of the quotient group $G / N$. However we don't know if $\text{Stab} (x)$ is necessary normal or not, since the Wikipedia page for "Group action" says that $\text{Stab}(x)$ is typically not a normal subgroup.
So how can we justify that $|G / \text{Stab}(x)| = [G: \text{Stab} (x)]$ if $\text{Stab}(x)$ happens not to be normal?
A subgroup has an index regardless of whether or not it is normal, and it still makes sense to talk about the sets of left or right cosets of a subgroup that is not normal. For example, a homogeneous space (a space that has a continuous group action sending any point to any other point) is the set of cosets of some subgroup of its homeomorphism group (this is virtually the same thing as the orbit-stabilizer theorem, just with some topology thrown in). In many contexts, normal subgroups are not interesting, such as in the symmetric group $S_n$ for $n\geq 5$. There are certain subgroups of the symmetric group, though, that have sets of cosets with a beautiful combinatorial structure (this is a finite Coxeter group thing). The moral of the story is the concept of cosets has numerous interesting applications that have nothing to do with a group structure on the quotient, and the orbit-stabilizer theorem is an example.