WDS: Allan Wechsler sent me various confused emails, but eventually unconfused himself, with the result his construction showing f(8)>=4, is now declared broken. My use of it to try to show f(n+6) >= 4+f(n) is similarly declared broken.
AW:: Taking a deep breath and trying again. I think my eight-direction configuration is not, in fact, realizable. The trouble is that AC, CE, EG, GA are all orthogonal pairs. I don't think that this spherical quadrilateral actually can flex in the way that we imagined. It can "fold", but one of the pairs AE, CG must remain antipodal. This forces identifications, so that the whole thing is reduced to the "bow-tie" configuration of five points we identified earlier, PQR+RST.
--WDS: Aha. The "flexible quadrilateral" ACEG... has a problem. If fix A at North pole of sphere of directions, then C,G both must be on equator no matter how you flex, which means E must be at either North or South pole no matter how you flex. Which means you lose, since A and E then represent the same direction, which is forbidden. It's kind of strange: essentially any spherical quadrilateral is flexible, but for this particular one, it has this miraculous sort of semi-rigid property. --However, to now make lemonade from lemons, this tells us that the 4-cycle is a forbidden subgraph, leading to stronger upper bounds. Specifically, an upper bound U on f(N) is: the most triangles any N-vertex graph G can have, where only graphs G not containing a 4-cycle, are allowed. On this topic, see http://arxiv.org/abs/1201.4912 which is about a different problem, but perhaps relevant.