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.
Best Answer
You don't deduce that $\int_{\partial\Omega} fv \cdot n\,ds = \int\int_\Omega \nabla \cdot (fv)\,dA$ you use it to make the desired conclusion. This fact is often known as the divergence theorem.
To spell it all out:
yes, $\partial \Omega$ is the boundary of $\Omega$.
The divergence theorem states that, if $V$ is a continuously differentiable vector field, then $$ \int_{\partial\Omega} V \cdot n\,ds = \iint_\Omega \nabla \cdot V\,dA $$
$\int\int_\Omega \nabla f \cdot v\,d{A} = \iint_{\Omega} \nabla \cdot (fv)\, dA - \int\int_\Omega f\nabla \cdot v\,dA$ as you've observed (basically by the 'product' rule for divergence).
By steps 2 and 3, $\iint_\Omega \nabla f \cdot v\,d{A} = \int_{\partial\Omega} fv \cdot n \, ds - \int\int_\Omega f\nabla \cdot v\,dA$
Now, if we let $v = \nabla g$ for some function $g$, we have $$\int\int_\Omega \nabla f \cdot \nabla g \,d{A} = \int_{\partial\Omega} f \nabla g \cdot n \, ds - \int\int_\Omega f (\nabla \cdot \nabla g)\,dA$$ i.e. $$\int\int_\Omega \nabla f \cdot \nabla g \,d{A} + \int\int_\Omega f (\Delta g)\,dA= \int_{\partial\Omega} f \nabla g \cdot n \, ds $$ which is Green's first identity.
To derive Green's second identity, write Green's first identity again, with the roles of $f$ and $g$ exchanged, and then take the difference of the two equations.