I'm trying to understand how do you get the Kelvin-Stokes theorem \begin{equation} \int_{S} (\nabla\times \omega) \cdot \mathrm{d}S = \int_{\partial S} \omega \cdot \mathrm{d}r \end{equation} from the generalized Stokes' theorem \begin{equation} \int_{M} \mathrm{d} \omega = \int_{\partial M} \omega \end{equation} I am aware that this question is answered here . However, I am looking at a more specific part of the question.
If I start looking at a vector $\omega = a \mathrm{d}x + b \mathrm{d}y + c \mathrm{d}z$, and take it's exterior derivative $\mathrm{d}\omega = \left(\partial_x b - \partial_y a \right) \mathrm{d}x \wedge \mathrm{d}y + \left(\partial_z c - \partial_y b \right) \mathrm{d}x \wedge \mathrm{d}y +\left(\partial_x c - \partial_z a \right) \mathrm{d}x \wedge \mathrm{d}z$ and notice that in the basis $\{ \mathrm{d}x^i \wedge \mathrm{d} x^j \}$ this is the cross product. But how to go from here - how to identify this with the inner product with $\mathrm{d}S$. How to even write what are the surface and line elements in general?
This is basically a question of how one converts back and forth between vector calculus and differential forms notation.
I suggest using an intermediary notation--the notation of clifford algebra and geometric calculus--which has the simplicity of vector calculus notation and the power of differential forms operations.
Let's do it starting with Kelvin-Stokes and working to writing it in the form of the generalized fundamental theorem of calculus ("Stokes' theorem").
First, in clifford algebra notation, the curl is written as "wedge derivative" instead:
$$\nabla \times \omega = i^{-1} (\nabla \wedge \omega) = \star (d\omega)$$
$i$ is the unit, right-handed pseudoscalar. Multiplying with $i$ (or its inverse) performs Hodge duality.
In vector calculus, when we write $F \cdot dS$ in a surface integral, we really mean that the surface is paramterized in some arbitrary way, with a normal vector $n$ such that $F \cdot dS = F \cdot n \, |dS|$.
In clifford algebra, we have a bivector $T$ that is Tangent to the surface. The relationship between $T$ and $n$ is that $n = Ti^{-1} = \star T$. Hence, we can write
$$F \cdot n = F \cdot (Ti) = (F \wedge T) i^{-1} = \star (F \wedge T)$$
That means the surface integral can be written as
$$\int_S (\nabla \times \omega) \cdot n \, |dS| = \int_S \{[i^{-1} (\nabla \wedge \omega) ]\wedge T \}i^{-i} \, |dS| = -\int_S (\nabla \wedge \omega) \cdot T \, |dS|$$
The minus sign comes from $i^{-1} i^{-1} = -1$.
Now, the other side of the identity can similarly be written
$$\int_{\partial S} \omega \cdot t \, |dr|$$
where $t$ is the tangent vector. The clifford algebra way of writing these integrals should be appealing: each involves a tangent $k$-vector. The tangent $1$-vector $t$ for a line integral and the tangent 2-vector $T$ for the surface integral.
Now, here is the missing piece in converting this to differential forms parlance: in differential forms notation , the tangent $k$-vector is implicit. It's not explicitly written down. Hence, the differential forms convention is that
$$\int_{\partial S} \omega \cdot t |dr| \equiv \int_{\partial S} \omega, \quad -\int_S (\nabla \wedge \omega) \cdot T \, |dS| \equiv \int_S d\omega$$
(The minus sign is not an issue here; bivectors square to negative scalars in clifford algebra. This merely reflects that, so that if the curl is aligned exactly with the surface, you get a positive answer.)
There's an additional wrinkle here: a differential forms integral written in these ways, according to this convention, must evaluate to a scalar--a number. But integrals like
$$\int_{\partial S} \omega \wedge t \, |dr|$$
couldn't be written under this convention because they don't evaluate to scalars. There are good reasons for this: such a quantity might have an unclear meaning on an arbitrary manifold, as you'd have to add bivectors from different tangent spaces. But vector calculus and clifford algebra's calculus can still write such an integral when in a flat vector space and make it a reasonable result.
Further, the choice of the tangent 2-vector $T$ and the tangent 1-vector $t$ is arbitrary. Any orientable manifold has two opposite orientations that can be used for an integral, and in differential forms, this is often related to the choice of parameterization by convention--a right-handed convention, say. In clifford algebra, we can choose either orientation for $T$ and the corresponding orientation for $t$ and get consistent results.