f(n) is

  int result = 0;
  for (int i = 1; i <= n; i *= 2)
     for (int j = 0; j <= i; j++)
        result++;

which can be simplified to

  int result = 0;
  for (int i = 1; i <= n; i *= 2)
     result += i+1;

which gives

  sum j=0..⌊log2 n⌋ of 2^j+1  =
  (sum j=0..⌊log2 n⌋ of 2^j) + ⌊log2 n⌋ + 1  =
  2^(⌊log2 n⌋+1) − 1 + ⌊log2 n⌋ + 1 =
  2 · 2^⌊log2 n⌋ + ⌊log2 n⌋

  n + log2 n − 1  ≤  f(n)  ≤  2n + log2 n

so is O(n)