Get specified bit in addition of two large binary numbers
Your method will indeed work, but for two $n$-bit numbers it will still take $O(n)$ time. There is a much faster way to solve your problem, with running time $\Theta(\log n)$, known as carry-lookahead addition. The downside is that in my experience teaching this technique, people have a hard time wrapping their heads around it. You can find online sources, here and here, and some algorithm or computer organization texts cover it. It's a really interesting algorithm, well worth taking the time to figure it out.