Residue theorem and combinatoric identities 
As highlighted, I don't quite follow the link from the contour integral to where summation terms are introduced and finally to the combinatoric identity although I do understand why the last step holds("since..")~
 A: It is just expansions of $(1+\epsilon)^n$
$$(1+\epsilon)^n=\sum_{k=0}^n\binom{n}{k}\epsilon^k$$
$$(1+\frac{1}{\epsilon})^n=\sum_{l=0}^n\binom{n}{l}\epsilon^{-l}$$
Multiply the two together:
$$(1+\epsilon)^n \cdot (1+\frac{1}{\epsilon})^n=\bigg(\sum_{k=0}^n\binom{n}{k}\epsilon^k\bigg)\cdot \bigg(\sum_{l=0}^n\binom{n}{l}\epsilon^{-l}\bigg)$$
$$=\sum_{k=0}^n\sum_{l=0}^n\binom{n}{k}\binom{n}{l}\epsilon^{k-l}$$
A: It's a basic fact from complex analysis that if $C$ is a closed contour around $a$, then 
$\frac{1}{2 \pi i} \int_{C} (z-a)^{k} dz=\left\{\begin{array}{ll}
1 & k=-1. \\
0 & \mbox{otherwise.}
\end{array}
\right.
$
This is discussed in many introductory textbooks on complex analysis but deriving it would take far more space than is reasonable for this site. 
If you're willing to accept the Cauchy formula stated above, then the rest of the derivation isn't too hard.
$\frac{1}{2 \pi i} \int_{C} \frac{1}{z} \sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} z^{k-l} dz=
\sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} \frac{1}{2 \pi i} \int_{C} z^{k-l-1} dz$
Let 
$\delta_{k-l}=\left\{ \begin{array}{ll}
1 & k-l=0. \\
0 & k-l \neq 0.
\end{array}
\right.
$
$z^{k-l-1}$ will be $z^{-1}$ only when $k=l$.  
Thus
$\frac{1}{2 \pi i} \int_{C} \frac{1}{z} \sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} z^{k-l} dz=
\sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} \delta_{k-l}.  
$
$\frac{1}{2 \pi i} \int_{C} \frac{1}{z} \sum_{k=0}^{n} \sum_{l=0}^{n} \binom{n}{k} \binom{n}{l} z^{k-l} dz=
\sum_{k=0}^{n} \binom{n}{k}^{2}.  
$
Note that there's an error in the source that you've quoted- since $k$ is the index of summation on the previous line, the next line can't simply be $\binom{n}{k}^{2}$.  
