Green's theorem in the plane is a special case of Stokes' theorem.
If we express Green's theorem as
$\oint_C M dx + N dy = \iint_R (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} ) dx dy$
we can express this in vector notation as
$Mdx + Ndy = (M \mathbf{i} + N \mathbf{j})(dx\mathbf{i}+dy\mathbf{j})= \mathbf{A}\cdot d\mathbf{r}$ in which $\mathbf{A} = M\mathbf{i}+ N\mathbf{j}$, $\mathbf{r} = x\mathbf{i}+y\mathbf{j}$.
We have
$\nabla \times \mathbf{A} = -\frac{\partial N}{\partial z}\mathbf{i}+\frac{\partial M}{\partial z}\mathbf{j}+ (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})\mathbf{k}$
and then $(\nabla \times \mathbf{A})\cdot \mathbf{k} = \frac{\partial N}{\partial x} -\frac{\partial M}{\partial y}$
So now we can re-write Green's theorem as (see * below re "dot" question)
$\oint_C \mathbf{A} \cdot d\mathbf{r} = \iint_R (\nabla x \mathbf{A})\cdot \mathbf{k}$ dR
in which dR = dx dy.
This is basically a problem from Shaum's Vector Analysis (Spiegel)
Edit: the full generalization of
$\oint_C \mathbf{A} \cdot d\mathbf{r} = \iint_R (\nabla x \mathbf{A})\cdot \mathbf{k}$ dR
to the usual version of Stokes' theorem
$\oint_C \mathbf{A} \cdot d\mathbf{r} = \iint_S (\nabla x \mathbf{A})\cdot \mathbf{n}$ dS
is also a problem in Shaum's, which is really an exercise in extending Green's to three dimensions. As Nunoxic points out above, a slightly different treatment of Green's theorem generalizes to Gauss' divergence theorem, also known as Green's theorem in space.
(*) "Dot" question: In your first equation, "curl v" is the result of dotting with the unit normal. I think the usage "curl" is confusing because, comparing your two equations, it appears that you have done something in $R^3$ that you did not do in $R^2$. That is not the case.
Let $\Omega$ be an open subset of $\mathbb{R}^n$ with $\partial\Omega$ of class $\mathscr{C}^\infty$, and let $X$ be a smooth vector field on $\Omega$. Now we compute
\begin{align}
d(i_X\operatorname{vol}_{\Omega}) & = d\left(i_X\left(dx^1\wedge\cdots\wedge dx^n\right)\right) \\
& = d\left(\sum_{i=1}^{n}(-1)^{i+1}X_i~dx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n\right) \\
& = \sum_{i=1}^{n}\frac{\partial X_i}{\partial x^i}~dx^1\wedge\cdots\wedge dx^n \\
& = (\operatorname{div}X)\operatorname{vol}_{\Omega},
\end{align}
where $\operatorname{vol}_\Omega$ denotes the volume form on $\Omega$, and $i_X$ denotes the interior product with $X$. From Stokes' theorem we obtain
$$\int_{\Omega}\operatorname{div}X~\operatorname{vol}_{\Omega}=\int_{\Omega}d(i_X\operatorname{vol}_{\Omega})=\int_{\partial\Omega}i_X\operatorname{vol}_{\Omega}.$$
Now decompose $X$ into it's tangential and normal components on $\partial\Omega$, i.e. $X=X^\top+X^\bot$. Then one easily computes
$$i_X\operatorname{vol}_{\Omega}=\operatorname{vol}_{\Omega}(X^\top+X^\bot,\cdots)=\operatorname{vol}_{\Omega}(\langle{X,\mathbf{n}}\rangle\mathbf{n},\cdots)=\langle X,\mathbf{n}\rangle \operatorname{vol}_{\partial\Omega},$$
where in the above $\mathbf{n}$ is the outward facing unit normal vector on $\partial\Omega$.
Now the usual divergence theorem follows immediately:
$$\int_{\Omega}\operatorname{div}X~\operatorname{vol}_{\Omega}=\int_{\partial\Omega}\langle X,\mathbf{n}\rangle \operatorname{vol}_{\partial\Omega}.$$
Best Answer
This may be a bit late but if k=1, we have that $\omega$ is a $0$-form scalar function $f$. It follows that $d\omega$ is a $1$-form. Hence, S is a 1-manifold curve C, while $\partial$$S$ is simply the two terminal points of the curve.
Now if you define a 1-manifold curve X($t$) and let $\omega$=$f(x)$ in $R^n$, and simply apply the Generalized Stokes Theorem, you can prove the following result:
$\int_C$ ($\nabla$$f$$\bullet$ $\overrightarrow T$)$ds$=$f(B)-f(A)$, where $\overrightarrow T$ is a (unit) tangent vector to the curve. Also, A and B are the endpoints of the curve.
Hint: Note that $\int_A^B$$f$=$f(B)-f(A)$. The difference in signs is due to the orientation of the endpoints of the curve itself.
For k=1 in $R^2$, ($\omega$ is a $0$-form), you simply get $\int_C$ ($\nabla$$f$$\bullet$ $\overrightarrow T$)$ds$=$f(B)-f(A)$ again, but $\nabla$$f$ only has 2 components.
For k=1 in $R$, the Generalized Stokes Theorem implies the Fundamental Theorem of Calculus. You can prove this rigorously or simply note that $(\nabla$$f$$\bullet$ $\overrightarrow T$)$ds$=$f ' (x) dx$ in $R$.
In $R^2$, the Generalized Stokes Theorem implies formulation A when working with k=2 ($\omega$ is a $1$-form). Note that formulation A (Green's Theorem) is precisely the Kelvin-Stokes Theorem in $R^2$. Also, note that Formulation B is equivalent to the divergence theorem in $R^2$.