Using 4-SUM to improve Subset Sum $O(2^{n/2})$ runtime
Work through the math; you are off by a factor of two.  Also, you might be confusing yourself by using the same letter $n$ for two different things.

Suppose we have a list of $N$ numbers.  You are proposing to split that into four chunks, where each chunk has $N/4$ numbers.  Enumerating all of the subsets of a single chunk takes $O(2^{N/4})$ time.  Now, you have a 4-sum problem where you are working 4 lists each containing $M=2^{N/4}$ items.  This 4-sum problem can be solved in $O(M^{k/2}) = O(M^{4/2}) = O(M^2)$, and $M^2 = 2^{N/2}$, so your proposed algorithm runs in $O(2^{N/2})$ time.

There is an algorithm for subset-sum that runs in time $O(2^{N/2})$ and space $O(2^{N/4})$.  The algorithm is due to Schroeppel and Shamir.  However, it's running time is still $O(2^{N/2})$, not $O(2^{N/4})$.

I don't think there is any known subset-sum algorithm whose running time is non-trivially better than $O(2^{N/2})$.  Blum, Kalai, and Wasserman have solved a related problem in subexponential time, but it's a different problem.