Merge sort: sorting and merging complexity $\Theta(n)$
Each iteration of merge sort consist of 2 phases:


Merge Sorting the first and the second half separately.
Merging the two halves.


So in your equation phase 1 is represented by $2T(n/2)$. This means that merge sort is called on the two halves. This is a recursive call, which is why $T$ is used here.

Phase 2 is represented by $\Theta(n)$. Merging two lists of length $a$ and length $b$ takes $\Theta(a+b)$. In our case the two lists are both of size $\frac{1}{2}n$, so we get a total of $\Theta(n)$.

In the end we get that $T(n) = $ step 1 + step 2, resulting in
$T(n) = 2T(n/2) + \Theta(n)$