It would appear based on small computations that for fixed $s$ the maximum occurs for $k$ roughly $\frac{s^2}{3}.$ Perhaps that maximum is very roughly $\frac{1}{s^2\sqrt{s}}.$

Here are the optimal $k$ values up to $s=20$: $$[1, 1], [2, 1], [3, 1], [4, 1], [5, 3], [6, 6], [7, 9], [8, 14], [9, 18], [10, 24], [11, 30]$$$$ [12, 37], [13, 45], [14, 53], [15, 62], [16, 72], [17, 82], [18, 93], [19, 105], [20, 118]$$
For $12 \le s \le 60$ the optimal $k$ is the integer closest to $0.3531s^2-1.236s+1.13.$ The latter value is sometimes above and sometimes below the rounded value. I started at $s=12$ just because that is the first time that $\frac{s^2}{4}$ is not enough. The only reason I stop at $60$ is because that is as far as I looked. A slightly better fit can be achieved by using more decimal places and/or starting the range higher. This is just a poor computational result, but perhaps it tells one to try to prove that $0.4s^2$ is too much and $0.25s^2$ is not enough for $s \ge 12$.

With a slight adjustment, the rounded quadratic fit for the optimal $k$ holds at least as far as $s=100$ at which point $f_k(s) \approx s^{-2.386}.$ The appropriate exponents at $s=25,50$ and $75$ are $-2.533,-2.449$ and $-2.41$