electrostatics - 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.