I don't have an answer to your main question but I can speak to some of your other points.

Firstly, Alperin diagrams are not so much graphs as lattices.  They are in fact an ordering on the composition factors of $M$ such that $N_1 \ge N_2$ iff there is a submodule with head $N_1$ and socle $N_2$. As such, triangles (as @AlexDugas [notes](https://mathoverflow.net/questions/327708/bounds-for-the-number-of-edges-in-an-alperin-diagram#comment818285_327708)) are not possible — $N_1 \ge N_2 \ge N_3$ and $N_3 \ge N_1$ are incompatible.

Secondly, a very important restriction on $M$ that must hold before we can construct Alperin diagrams is that $M$ must be multiplicity free.  That is, each factor must appear in a composition series exactly once.

To see why this is important, consider two identical modules with composition factors $[N_1, N_2]$ (so the head is $N_1$).  Then the direct sum of these modules, quotiented out by a suitable diagonal image of $N_2$ will have head $N_1\oplus N_1$ and socle $N_2$.  What is its Alperin diagram?  At first we may say
```
N_1   N_1
   \ /
   N_2
```
but in fact a suitable change of basis gives that $N_1$ falls out as a direct summand!  Thus a "better" diagram is the forest
```
         N_1
N_1  +    |
         N_2
```
Which should we use?  This is a simple case, but you can imagine that in a module with a dozen factors some of which are repeated multiple times, this becomes a serious question.