Kolmogorov complexity of a sequence of n bits with k ones
In Kolmogorov complexity, our measure of complexity is program length. We don't care about the resource consumption of the program (unless we're interested in resource-bounded Kolmogorov complexity). Given $n$ and $k$, a program can generate all $n$-bit strings of weight $k$, so there is no need to store the list itself as part of the program.

The formula you quote and the explanation you give seem mistaken. In order to use the program you mention to generate a specific vector, we need to give it as input $n,k,i$. It takes $K(n) \le \log n + O(\log\log n)$ bits to encode $n$ itself. Given $n$, we can encode $k$ in $\lceil \log n \rceil$ bits, and $i$ in $\lceil \log \binom{n}{k} \rceil$ bits. Overall, we obtain that the Kolmogorov complexity is at most
$$ 2\log n + \log \binom{n}{k} + O(\log\log n). $$