Big-Oh of recursive function
The reason is easier to see when you figure the big-$\mathrm O$ of the loop and the recursion (last line) separately and then add them. The loop takes $\mathrm O(n^2)$ time, and the recursive part takes twice the time of $f(n/2)$.

So we get 
$$\mathrm{time} (f(n))=\mathrm O(n^2 + 2\mathrm{time}(f(n/2)))\tag a$$

and we can simplify this  by noticing that this will form a geometric sequence.

$$ = \mathrm O(n^2+2\left(n/2\right)^2 + 4\mathrm{time}(f(n/4)))\tag b$$
$$ = \mathrm O(n^2 + 2(n/2)^2 + 4(n/4)^2 + 8\mathrm{time}(f(n/8)))\tag c$$
$$= \mathrm O\left(\sum_{i=0}^{\inf} \frac{n^2}{2^{i}}\right)\tag d$$
$$= \mathrm O\left( n^2\sum_{i=0}^{\inf}\left(\frac{1}{2^{i}}\right)\right)\tag e$$

$\sum_{i=0}^{\inf}\left(\frac{1}{2^{i}}\right)$ on line $\mathrm (e)$ converges to a  nonzero value that is not dependant on $n$ (specifically, $2$), so it can be removed from the $\mathrm {O}()$, to give just $\mathrm O(n^2)$



Note that this way of solving (leaving constant factors and ignoring the base case of recursion) will not work for all problems.

For example what if the inner loop had time complexity $\mathrm O(n)$? $$\mathrm{time}(f(n)) = \mathrm O(n + 2\mathrm{time}f(n/2)) = \mathrm O(n + 2\frac{n}2+4\mathrm{time}f(n/4)) = ...$$ which would give $\mathrm O\left( n^2\sum_{i=0}^{\inf}1\right)$, which is obviously not correct because it diverges.
However, if we ingore the constant factor of 2, we get 
$$\mathrm{time}(f(n)) = \mathrm O(n + \mathrm{time}f(n/2)) = \mathrm O(n + \frac{n}2+\mathrm{time}f(n/4)) = ...$$ which gives $\mathrm O\left( n^2\sum_{i=0}^{\inf}\frac1{2^n}\right)$, which converges to $2$

-- End note