Each line segment with ends from the set $S$ of length $\frac{k}{n}$ (not lying on a side of the triangle) is the diagonal of one and only one rhombus of type $M$. The segments parallel to one side of the triangle are
$$1 + 2 + 3 + \ldots + (n-1) = \frac{n(n-1)}{2}.$$
Therefore, the total number of rhombi of type $M$ is:
$$m = 3 \frac{n(n-1)}{2}.$$

For the counting of rhombi of type $D$, we distinguish the points of $S$ which are interior points of the triangle $ABC$ into three categories as follows.

The first category of points are centers of exactly one rhombus of type $D$. These points are only 3, for every $n$. In figure (3) you can see the case for $n = 8$.

The second category consists of points which are centers of exactly two rhombi of type $D$. These points lie on the segments which are parallel to the sides of the triangle $ABC$ and at the shortest possible distance from them. On each such segment there exist $n-4$ such points, and therefore we have $3(n-4)$ points of this category. In figure (4) you can see these points for $n = 8$.

The third category consists of the rest of the points which are centers of three rhombi of type $D$. These points are totally:
$$1 + 2 + \ldots + (n-5) = \frac{(n-5)(n-4)}{2}.$$
In figure (5) you can see these points for $n = 8$.

Hence, the number of rhombi of type $D$ is the following:
$$d = 3 + 3(n-4) + 3 \frac{(n-5)(n-4)}{2} \Leftrightarrow d = \frac{3}{2}[2 + (n-1)(n-4)].$$

Finally, we have $m - d = 3(2n - 3)$.