Explain the Stokes -theorem from differential from into Integral form 
I want to understand the Stokes -theorems deeper. I am trying to understand the operation from 
$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$$ 
to 
$$\oint_{\partial \Sigma} \mathbf{E} \cdot d\boldsymbol{\ell} = - \int_{\Sigma} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}$$ 
relating to Maxwel-Faraday's law here.

You have there surface-integral, closed line-integral, curl and the rate of change. Explain the transition from the differential from into the integral form.
Bonus points and Puzzles

  
*
  
*Notation? I am unsure whether $\int$ is just physical convention, instead of $\int\int$ for the surface integral like the Kelvin-Stokes -theorem $\oint_{\Gamma} \mathbf{F}\, d\Gamma 
 = \iint_{\mathbb{S}} \nabla\times\mathbf{F}\, d\mathbb{S}$, the two forms of notation confuses me greatly -- mathematical notation and physical notation?
  

 A: You need to use the theories below. The notation with integrals between the two cases in KS -threorem look different: $\int\int_S$ and $\int_\Sigma$. I think they are still meaning the same thing. The $\partial\Sigma$ means the boundary of the manifold $\Sigma$.
I am unable to get inside this theorem so I am unable to explain how to get from the differential from into the integral form, anyway below some working, perhaps someone could continue here.
Trials

I. Suppose I integrate $\nabla \times \mathbf{E} = -\frac{\partial
 \mathbf{B}} {\partial t}$ with respect to the area so 
$$\int_\Sigma\left(\nabla\times \bf E \right)\cdot d\bf A=-\int_\Sigma
 \frac{\partial \mathbf{B}} {\partial t}\cdot d\bf{A}:=RHS.$$
II. Suppose I integrate it with respect to the line so
$$\oint_{\partial \Sigma} \left(\nabla\times \mathbf{E}\right) \cdot
 d\boldsymbol{\ell} = \oint_{\partial \Sigma} \left(-\frac{\partial
 \mathbf{B}} {\partial t}\right) \cdot d\boldsymbol{\ell}$$ 
err no LHS emerging.
III. Some other way? I am clearly not getting sides equal like that, at least easily.

Theories

  
*
  
*Stokes theorem $\int_{\partial \Omega} w =\int_\Omega dw$ (sthing about boundaries and manifolds
  here).
  
*Kelvin-Stokes -theorem  $$\oint_{\Gamma} \mathbf{F}\, d\Gamma 
 = \iint_{\mathbb{S}} \nabla\times\mathbf{F}\, d\mathbb{S}$$ (source)
  or  $$\int_{\Sigma} \nabla \times \mathbf{F} \cdot d\mathbf{\Sigma} =
 \oint_{\partial\Sigma} \mathbf{F} \cdot d \mathbf{r}$$
  (source).

Perhaps related here.
A: If I undertstand your query correctly, all you really need to complete the thought are Stoke's and Gauss' Theorems:
$$ \iint_{\partial E} \vec{F} \cdot d\vec{S} = \iiint_E \nabla \cdot \vec{F} dV $$
$$ \int_{\partial M} \vec{F} \cdot d\vec{r} = \iint_{M} \nabla \times \vec{F} \cdot d\vec{S}$$
Here we suppose that $E$ is a simple solid region with boundary $\partial E$ and $M$ is a simply connected surface with boundary $\partial M$. The boundaries must be consistent with the interiors of the integration regions. This means $\partial E$ has outward pointing normal whereas a trip around $\partial M$ finds the interior of $M$ always on the left of your journey.
Begin with $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$ and integrate over some surface $M$ at time $t$. Apply Stokes' theorem to convert the flux integral of the curl to the line integral of the boundary:
$$ \iint_{M} -\frac{\partial \vec{B}}{\partial t} \cdot d\vec{S}=\iint_{M} \nabla \times \vec{E} \cdot d\vec{S}=\int_{\partial M} \vec{E} \cdot d\vec{r} = \mathcal{E}_{induced}$$
Moreover, the term on the l.h.s. of the above can be expressed as $\frac{\partial}{\partial t} \iint_{M} \vec{B} \cdot d\vec{S} =  \partial_t \Phi_B$. We thus derive the integral form of Faraday's Law; the voltage induced around a closed loop is proportional to the change in the magnetic flux through the loop.
