Solving the recurrence relation T(n) = 2T(n/2) + nlog n via summation
First, you can get the the result from the second xase of master theorem. Second, you have a mistake in your exapnsion:

$$T(n) = 2T(\frac{n}{2}) + n\log(n) = 2(2T(\frac{n}{2^2} + \frac{n}{2}\log(\frac{n}{2})) + n\log(n) = n\log(n) + n\log(\frac{n}{2}) + 2^2T(\frac{n}{2^2}) =  n\log(n) + n\log(\frac{n}{2}) + 2^2(2T(\frac{n}{2^3} + \frac{n}{2^2}\log(\frac{n}{2^2})) = n\log(n) + n\log(\frac{n}{2}) + n \log(\frac{n}{2^2}) + 2^3T(\frac{n}{2^3})$$

Hence, we will have the following, if we suppose $T(1) = 0$, w.l.o.g:

$$T(n) = \sum_{i=0}^{\log(n)}n\log(\frac{n}{2^i}) = n \log(\Pi_{i=0}^{\log(n)}\frac{n}{2^i}) = n \log(\frac{n^{\log{n}}}{2^{\frac{\log(n) \times (\log(n) + 1)}{2}}}) = n \log(n) \log(\frac{n}{\sqrt{2}\sqrt{n}}) = n \log(n)\log{\frac{\sqrt{n}}{\sqrt{2}}} = \Theta(n
\log^2(n))$$

Be aware that, if we change the values of $T(1)$ and $T(2)$ to the given constant values, nothing changed in the asymptotic analysis.