It's not O(n^2) ya doofus. It's O(n * k), where n is number of splits, and k is the select (or vice versa). It's still very time complex, but it doesn't grow by the number of splits squared.

It's just named "sorted", not that it's sorted

Please use spoiler flag next time, your post was visible in the homepage.

As I understand it, it's just how BitConverter works. It seems that whatever architecture CodeWars uses to run our code happens to be little-endian, and thus GetBytes() acts accordingly. See the documentation:

The order of bytes in the array returned by the GetBytes method depends on whether the computer architecture is little-endian or big-endian.

Do note that the Kata does implicitly describe the first octet being the left-most one (and, correspondingly, the most significant one).