You are using Gauss's Law. $\ Electric\ Flield's\ Flux= \frac{Q}{\epsilon_0}$
Calculate the flux through the 6 surfaces first.
When $\Delta x\rightarrow0$, Then the contribution of flux coming from the two surfaces that are shaded surfaces as well as the other two "non- radial" surfaces becomes negligibly small. You can ignore them in the limit. I hope you are following. Only the 2 big surfaces where $\ df$ is marked will contribute because the limit does nothing to their area.
Now simply calculate the flux from these two. It is
$\ df\ \vec{n}\cdot\vec{E_a}$ for one and $\ -df\ \vec{n}\cdot\vec{E_i}$ for the other. Negative sign coming in the second case because the flux is inward. Net charge lies on the surface only, so $\ Q=df \ \sigma$.
Use Gauss's Law. $\ df$ will cancel giving you what you want
