@PietroMajer No I think, for example for $k=2$, we have $f(k)=9$, but your parameter is $6$. Your parameter is exactly $3k$, isn't it?

@JosephO'Rourke The sketch of proof for f(2)=9 is using some cases for those nine points, 5+4 or 6+3 or..., then replacing one of them with the new point (10th point). The counterexample for 8 is three circles with their pairwise intersections (6 points) and one extra point on each of them. I think the answer for "unit circles" is very small.

@PietroMajer No, the number $f(k)−1$ is the maximum n such that there exist a set $S$ of points $P_1,P_2,...,P_n$ such that for each $i$, $S−P_i$ is on $k$ circles but the whole $S$ is not.

@Gerry Myerson Good example for 3D, but Of course planarity is important, if not you can find infinitely many three colored triangles in the 3 dimensional space. ($1\times1\times\infty$)