How general formulation of Stoke's theorem relate to Kelvin-Stokes theorem I asked a similar question, but I realized the question is too vague and it's better to start a new one:
We know that there are two usually used formulations of Stoke's theorem. One is vector calculus's usage of Stoke's theorem called Kelvin-Stokes theorem. And one is Stoke's theorem that involves manifolds and boundary of mainfolds and differential forms.
The question is, how does one derive Kelvin-Stokes theorem from the general manifold formulation of Stoke's theorem?
 A: You can recapture the Kelvin-Stokes Theorem from the General Stokes Theorem in the following manner:
Let $\mathbf{F} = 
\left( \begin{array}{ccc}
\mathbf{F}_1 \\
\mathbf{F}_2 \\
\mathbf{F}_3 
\end{array} \right)$ be a vector field in $\mathbb{R}^3$, and let $\mathcal{M}$ be a Manifold. Define the differential 1-form $\omega$ as $\displaystyle{ \omega = \sum_{i=1}^3 {\mathbf{F}_i \mathrm{d} x_i} }$. Then, $\displaystyle{ \mathrm{d} \omega = \sum_{i=1}^3 {\mathrm{d} \mathbf{F}_i \wedge \mathrm{d} x_i} }$, which by the General Stokes Theorem
$$
\int \limits_\mathcal{\partial M} {\omega} = \int \limits_\mathcal{M} {\mathrm{d} \omega}
$$
gives,
$$
\int \limits_\mathcal{\partial M} {\sum_{i=1}^3 {\mathbf{F}_i \mathrm{d} x_i}} = \int \limits_\mathcal{M} {\sum_{i=1}^3 {\mathrm{d} \mathbf{F}_i \wedge \mathrm{d} x_i}}.
$$
Now, define $\mathrm{d}\mathbf{r}$ as $\left< \mathrm{d} x_1, \mathrm{d} x_2, \mathrm{d}x_3 \right>$; then, 
$$
\int \limits_\mathcal{\partial M} {\omega} = \int \limits_\mathcal{\partial M} {\sum_{i=1}^3 {\mathbf{F}_i \mathrm{d} x_i}} = \int \limits_\mathcal{\partial M} {\mathbf{F} \cdot \mathrm{d} \mathbf{r}} .
$$
Furthermore, 
\begin{align}
\int \limits_\mathcal{M} {\sum_{i=1}^3 {\mathrm{d} \mathbf{F}_i \wedge \mathrm{d} x_i}} &= \int \limits_\mathcal{M} {\sum_{i=1}^3 {\left( \frac{ \partial \mathbf{F}_i }{ \partial x_1 } \mathrm{d} x_1 + \frac{ \partial \mathbf{F}_i }{ \partial x_2 } \mathrm{d} x_2 + \frac{ \partial \mathbf{F}_i }{ \partial x_3 } \mathrm{d} x_3 \right) \wedge \mathrm{d} x_i}} \\
%
%
& \vdots \\
%
%
&= \int \limits_\mathcal{M} { \left( \frac{\partial \mathbf{F}_3 }{ \partial x_2 } - \frac{\partial \mathbf{F}_2 }{ \partial x_3 } \right) \mathrm{d} x_2 \wedge \mathrm{d} x_3 + \left( \frac{\partial \mathbf{F}_1 }{ \partial x_3 } - \frac{\partial \mathbf{F}_3 }{ \partial x_1 } \right) \mathrm{d} x_3 \wedge \mathrm{d} x_1} \\
%
%
& \hspace{4 mm} + {\left( \frac{\partial \mathbf{F}_2 }{ \partial x_1 } - \frac{\partial \mathbf{F}_1 }{ \partial x_2 } \right) \mathrm{d} x_1 \wedge \mathrm{d} x_2 } \\
&= \int \limits_\mathcal{M} { \left( \nabla \times \mathbf{F} \right) \cdot \mathrm{d} \left< x_2 x_3, x_3 x_1, x_1 x_2 \right> } = \int \limits_\mathcal{M} { \left( \nabla \times \mathbf{F} \right) \cdot \mathrm{d} \mathcal{M} }.
\end{align}
and thus,
$$
\int \limits_\mathcal{\partial M} {\mathbf{F} \cdot \mathrm{d} \mathbf{r}} = \int \limits_\mathcal{\partial M} {\omega} = \int \limits_\mathcal{M} {\mathrm{d} \omega} = \int \limits_\mathcal{M} {\sum_{i=1}^3 {\mathrm{d} \mathbf{F}_i \wedge \mathrm{d} x_i}} = \int \limits_\mathcal{M} { \left( \nabla \times \mathbf{F} \right) \cdot \mathrm{d} \mathcal{M} } .
$$
