You can compute the average very explicitly. For $v\in F_2^N$, let $||v||$ denote the Hamming weight of $v$. Changing the order of summation, your sum can be re-written as
$$ \sum_{v\colon\ ||v||=k} T(v), $$
where $T(v)$ is the number of $K$-dimensional subspaces of $F_2^N$, containing $v$. Now, if $k>0$, then $v\ne 0$, and the number of subspaces in question does not depend on $v$ and is equal to the number of $(K-1)$-dimensional subspaces of $F_2^{N-1}$, which is
$$ \frac{(2^{N-1}-1)(2^{N-1}-2))(2^{N-1}-4)\cdots(2^{N-1}-2^{K-2})}{(2^{K-1}-1)(2^{K-1}-2)(2^{K-1}-4)\cdots(2^{K-1}-2^{K-2})}. $$
The sum under consideration is therefore equal to the product of this fraction and the binomial coefficient $\binom Nk$ (the number of vectors $v$ of weight $k$), and the average is obtained by dividing this sum by the total number of all $K$-dimensional subspaces of $F_2^N$, which is
$$ \frac{(2^N-1)(2^N-2)(2^N-4)\cdots(2^N-2^{K-1})}{(2^K-1)(2^K-2)(2^K-4)\cdots(2^K-2^{K-1})}. $$
After cancellations, the average is equal to
$$ \frac{2^K-1}{2^N-1}\,\binom Nk = 2^{K-N}\binom Nk \, (1+o(1)),\ K\to\infty. $$