Clarification about conclusion of $Θ(nlogn)$?
In the first part where the summation $\sum_{i=1}^{\log n}i2^i$, you see that whole summation can be written as that chain of summations (if you sum them all up you'll end up getting the first form on the top). Then you use the summation of power two's formula, i.e. $\sum_{i=0}^{n}2^i= 2^{n+1}-1$. 

In the first line $=2+2^2+...+2^{log n}$ the formula is missing a -1 so the result will be $2^{\log n +1}-2$, the second one is missing a 2 so it will be $2^{\log n+1}-2^2$ and so on.

You can write $2^{\log n +1}$ as $2*2^{\log n}$ which is equal to $2*n$. I think what confuses you is that you think they wrote $2^{\log (n+1)}$ where as it is $2^{\log n +1}$.

As for why the $2n*\log n - (2^{\log n +1} -2) \in \Theta(nlogn)$, since $2n\log n \in \Theta(nlogn)$ and $2^{\log n + 1} = 2*n$ thus it is $\Theta(n)$ and from the property of asyptotic notation we can conclude $\Theta(nlogn)-\Theta(n) \in \Theta(nlogn)$.