The discontinuity of Electric Field ''electric field always undergoes a discontinuity when you cross a surface charge $\sigma$'' GRIFFITHS
In the derivation; Suppose we draw a wafer-thin Gaussian Pillbox, extended just barely over the edge in each direction. Gauss law states that:
$$\int_{S} E \cdot A = Q_\text{enc}/ \epsilon$$
and so $$E_{\perp\;\text{above}} - E_{\perp\;\text{below} } = \sigma/ \epsilon .   $$
My question is why not $2A$? 
$$\int_{S} E \cdot A = 2EA$$ 
because the top area of pillbox and the bottom area of pillbox, just as because the 2 parts of the flux...
SO.. WHY NOT : $$E_{\perp\;\text{above}}- E_{\perp\;\text{below} } = \sigma/ 2\epsilon  ~?  $$
And why there is tangencial component of electric field; not just perpendicular to the surface, which can be seen as flat just looking very close to the surface.
 A: Immediately above the surface, write
$$
\mathbf{E}=E_{\text{above}}\widehat{\mathbf{n}}+\mathbf{E}_\text{tangential,above}
$$
and immediately below the surface, write
$$
\mathbf{E}=E_{\text{below}}\widehat{\mathbf{n}}+\mathbf{E}_{\text{tangential,below}},
$$
where $\widehat{\mathbf{n}}$ is the unit surface normal pointing in the above direction, and both tangential components, by definitition, are orthogonal to $\widehat{\mathbf{n}}$.  In the limit where the "pillbox" gets arbitrarily small, we can treat $E_{\text{above}}$ and $E_{\text{below}}$ as approximately constant in the pillbox, so that the flux is given by
$$
A(E_{\text{above}}-E_{\text{below}})+O(h),
$$
where $h$ is the height of the pillbox (this is Big $O$ notation).  The minus sign arises because the unit normal to the surface below is $-\widehat{\mathbf{n}}$, not just $\widehat{\mathbf{n}}$.  By Gauss's Law, this is equal to $\frac{A\sigma}{\varepsilon _0}$, and hence, in the limit
$$
\boxed{E_{\text{above}}-E_{\text{below}}=\frac{\sigma}{\varepsilon _0}}.
$$
Obviously, techincally speaking more careful limiting arguments are needed, but hopefully it is clear that this argument can be made completely precise.  Let me know if you have any more questions.
To be fair, this is essentially exactly what Ron was getting at, but hopefully the increased detail made this more clear.
A: Because one value of E, the inner one, multiplies the inner A, and the outer E multiplies the outer A, and you then subtract the two.
A: The tangential component due to the locally flat piece of surface, is indeed zero, at the surface. But, the total electric field is the field due to the locally flat patch, plus the rest of the surface. Hence, the total tangential component need not be zero. 
A: there will  be no tangential component of E^ in the case of the charged surface in electrostatics. If so then this would result in a force on the free electrons which would cause a drift hence a current which is not desired in electrostatics.
A: Maybe this will help:
http://physweb.bgu.ac.il/ARCHIVE/Courses/2011B/Electro1/extra/Discontinuity_in_an_Electric_Field.pdf
I think the issue is the way you define: $$E_{above}; E_{below}$$
Take a look at the link and note that, for example:
$$E_{above} = \vec{E}_2\cdot\hat{n}$$
