It means "open" the recursion.
For simplicity - denote $O(n)$ as $c\cdot n$:
T(n) &= 2T(n/2) + cn \\
&= 2(2T(n/4) + cn/2) + cn\\
&= 2 (2 (2T(n/8) + cn/4) + cn/2) + cn\\
It might give you intuition, but it is NOT a proof. To prove it, you will need mathematical induction or the master theorem.
Proving with induction (assuming
O(n) component is
n for simplicity):
T(n) <= n*logn + n
T(1) = 1 (assumption)
Assumption: the claim is correct for all
k < n.
T(n) = 2T(n/2) + n = (assumption) <= 2* (n/2 * log(n/2) + n/2) + n
= n*log(n/2) + 2n = n*(log(n)-log(2)) + 2n = (assuming base 2 for log)
= n*(log(n) -1 ) + 2n = nlogn -n + 2n = n*logn +n