Here the professor said that, Tournament sort needs (n-1) + 2(n-1)logn comparisons.
{Where (n-1) for calculating Maximum or say creating Tournament structure and 2(n-1)logn for other elements to sort}
Why did the professor leave out the number of comparisons needed to find the Minimum? Because to calculate min elements we need (n/2 - 1) comparisons.
Here logn means log n to the base 2
I am watching NPTEL lecture https://youtu.be/NU77XIX-wro (Video Time - 25:50)
Because of "sorting" and "leaving out" I think, you refer to the big O notation.
If the function f can be written as a finite sum of other functions, then the fastest growing one determines the order of f(n).
In (n-1) + 2(n-1)logn, n grows not as fast as n*log(n).
It looks like finding the minimum is not one of the steps of the algorithm. So the number of comparisons required in order to find the minimum is irrelevant, since that step is never even performed.
