Where did the Orbit-Stabilizer came from (not historically)? The orbit-stabilizer theorem is completely encoded by the equation
$$|G| = |\operatorname{Orb}(x)||\operatorname{Stab_G}(x)| $$
Most books/online presentations I am reading jump straight into this equation after the definitions are introduced. 
Note that Lagrange Theorem tells us 
$$|G| = [G: \operatorname{Stab}_G(x) ]|\operatorname{Stab}_G(x)|$$
So what prompts us to suggest $[G:\operatorname{Stab}_G(x)]$ is bijective with $|\operatorname{Orb}(x)|$?
Is it observed via a few examples and conjectured later? 
Note I am not asking for the proof
 A: I'm going to answer what I think is the literal question; if I've misunderstood the question, perhaps you can correct me.
I believe what you are asking is: How can a mathematician discover this theorem, knowing nothing about it beforehand?
Well, the answer is that they can't. That's not what happens. Just learning the abstract definition of a group, and maybe the definition of a group action, and maybe the definition of orbits, and then expecting that anyone can just say "Hey, here's a theorem!"... well... that's not how any mathematical theorems ever get discovered.
Instead, someone learns about actual groups, and actual group actions. They learn examples. They observe patterns. They stumble upon this particular pattern, noticing that it holds in a few different examples: the order of the group is the size of an orbit times the size if a stabilizer of a point in that orbit. They think "Huh... is this just a coincidence?" They might look for more examples to bolster the point, they might look (unsuccessfully) for counterexamples, which bolsters the point even more. 
They become more and more convinced that the pattern is true. 
And when a mathematician becomes convinced that something is true, then they are very motivated to prove that it is true.
And fortunately, the proof is easy.
I really don't think there's much more than that to say.
A: Here is one motivation. Call a set on which $G$ acts a "$G$-set". One particular example is that if $H$ is a subgroup, then the set $G / H$ of cosets of $H$ is a $G$-set in the obvious way: $g \cdot (xH) := gx H$. The "orbit-stabilizer" theorem is saying that this example is "canonical" in the sense that EVERY example is essentially this example.
What I mean is that another way of phrasing the theorem is that, viewing $\mathrm{Orb}(x)$ as a $G$-set, it is isomorphic to the coset space $G / \mathrm{Stab}(x)$. So every (transitive) $G$-set is equivalent to some coset space $G/H$. The version of the theorem that you mention is a corollary of this fact by taking the cardinality of both sets.
A: Hint: the mapping $p$ from $G$ to $X$ defined by $p(g) = gx$, partitions $G$ by the equivalence relation $\simeq$ where $g \simeq h$ iff $p(g) = p(h)$. This equivalence relation has $|\mathrm{Orb}(x)|$ equivalence classes and each equivalence class has $|\mathrm{Stab}(x)|$ elements.
A: Let's try to motivate the theorem this way.
If $G$ acts on $X$ then clearly the orbits are important because if two elements are in the same orbit, then they "look the same with respect to $G$", and if they are different orbits, then they are different with respect to $G$.
So when studying the action of $G$ on $X$, it makes sense to study one orbit at a time because two different orbits are unrelated as far as $G$ is concerned.
So we pick a point $x \in X$ and consider its orbit in $G$.  Now we ask ourselves, what can we say about this orbit? To say that $y$ is in the same orbit as $x$ means that there exists $g \in G$ such that $gx = y$.  So it is a reasonable question to ask, how many such elements $g$ are there?  From there, a little thought leads one to the conclusion that this number does not depend on $y$ (as long as it is in the orbit of $x$), because elements in the same orbit "look the same with respect to $G$". Thus we may take $y=x$, so we are studying elements such that $gx=x$, i.e., the stabilizer.
From here, it is not far to the orbit-stabilizer theorem.
I hope this was helpful.
