Call the circles which contain at least three of the points from $M$ interesting. The problem's conditions tell us that every triplet of the points belongs to exactly one interesting circle. Lemma. $\sum_{P\in M} f(P) = 0$. Proof. For a triplet of points $T = \{P, Q, R\}$ let $f(T) = f(P) + f(Q) + f(R)$. Consider $\sum_{T\subset M} f(T)$, with $T$ ranging over every possible triplet. On one hand, every point of $M$ is included in exactly $\binom{n-1}{2}$ triplets, so \[ \sum_{T\subset M} f(T) = \binom{n-1}{2}\sum_{P\in M} f(P). \] On the other hand, suppose we choose an interesting circle $\mathcal{C}$ and count all the triplets which are entirely on it. Then if the circle has $k$ points from $M$, each point will be included in $\binom{k-1}{2}$ triplets; thus \[ \sum_{T\subset M\cap\mathcal{C}} f(T) = \binom{k-1}{2} \sum_{P\in M\cap\mathcal{C}} f(P) = 0. \] Now we simply sum over all interesting circles; since each triplet is included in exactly one interesting circle we have \[ \sum_{T\subset M} f(T) = 0 \] and combining these two equations proves the lemma. Now let $A$ and $B$ be two arbitrary points and let $S$ be the set of all interesting circles which pass through both of them. Not all of the points of $M$ are concyclic, hence $|S| > 1$. Also, every point other than $A$ and $B$ which is in $M$ is in exactly one circle of $S$, so \[ 0 = \sum_{\mathcal{C}\in S}\sum_{P\in \mathcal{C}\cap M} f(P) = |S|(f(A) + f(B)) + \sum_{P\in M\setminus\{A, B\}} f(P). \] Applying the lemma, we have $0 = (|S|-1)(f(A) + f(B))$; since $|S| > 1$ we know $f(A) + f(B) = 0$. The rest of the proof is easy: for three arbitrary points $A, B, C$ we have $2f(A) = (f(A) + f(B)) + (f(A) + f(C)) - (f(B) + f(C)) = 0$ so the only function that works is the zero function.