For brevity, rewrite the recursion as$$f_k(n)=\sum_{1\leqslant i\leqslant\frac n2}\binom nif_k(i)f_k(n-i).$$Now divide it by $n!k^n$ and rewrite like this:$$\frac{f_k(n)}{n!k^n}=\frac1{n!}\sum_{1\leqslant i\leqslant\frac n2}\binom nii!(n-i)!\frac{f_k(i)}{i!k^i}\frac{f_k(n-i)}{(n-i)!k^{n-i}}.$$
It follows that for$$g_k(n):=\frac{f_k(n)}{n!k^n}$$we get$$g_k(n)=\sum_{1\leqslant i\leqslant\frac n2}g_k(i)g_k(n-i),$$and $g_k(1)=1$ (in particular $g_k(n)$ does not depend on $k$).

This is the recursion for A000992 (number of (unlabeled, rooted) ordered trees on $n-1$ vertices in which all outdegrees are $\leqslant2$ and, for each vertex of outdegree $2$, the sizes of its two subtrees are weakly increasing left to right ($n\geqslant2$)). That page contains some information about these numbers.