Proving Fundamental Theorem of Calculus given Stokes' Theorem

Accepting Stokes' theorem (101) as a premise, how do I prove the fundamental theorem of calculus (102)? I know that the FTC is a straightforward specialization of Stokes' theorem and so proving $$\text{(101)} \implies \text{(102)}$$ should be trivial. I'm using the triviality of this implication to try to improve my understanding of Stokes' theorem. I vaguely remember Stokes' theorem from school, but didn't really understand it then and don't really understand it now. I wrote both theorems with the boundary-related part on the left and the interior-related part on the right.

Basically, what values do I pick for $$\Omega$$ and $$\omega$$ to turn (101) into (102)?

$$\int_{\partial \Omega} \omega = \int_{\Omega} d \omega \tag{101}$$

$$F(b) - F(a) =\int_{a}^{b} f(x)dx \tag{102}$$

Explaining the notation, $$\omega$$ is a differential form and and $$\Omega$$ is an orientable manifold. I think this means that $$\Omega$$ is not inherently equipped with an orientation but is capable of receiving one. So, $$\Omega$$ cannot be something like a Möbius strip.

I'm trying to get the left sides to match first, but I'm stuck.

So it seems like the most straightforward way to prove the implication is to have $$\Omega$$ be a closed interval on the real line $$[a, b]$$.

However, the boundary of $$[a,b]$$ is a set of two points $$\{a, b\}$$. I'm trying to understand what $$\omega$$ would have to be to make $$\int_{\{a, b\}} \omega$$ equal to $$F(b)-F(a)$$ rather than $$F(b)+F(a)$$.

What values to pick for $$\Omega$$ and $$\omega$$?

You should take $$\omega = F$$ (which is a $$0$$-form), and you should take $$\Omega = [a,b]$$ as you say, with the natural orientation (given by choosing the identity charts to be positively oriented, or to say it a different way, by choosing the nonvanishing $$1$$-form $$dx$$ on $$[a,b]$$, where $$x$$ is the identity coordinate function). This induces an orientation on $$\partial[a,b] = \{a,b\}$$, giving $$\{a\}$$ negative local orientation and $$\{b\}$$ positive local orientation, so that integrating $$F$$ on it gives $$F(b)-F(a)$$ instead of $$F(b)+F(a)$$. The statement of Stokes' theorem includes the assumption that $$\partial \Omega$$ is also oriented, in this specific way induced by the orientation on $$\Omega$$. Look here for more details in the 1-dimensional case.