If a < b then a^2 < b^2. This is not true if you let a = 0 and b = h. The dual numbers do not have an ordering. You should provide sources and proofs next time because it seems like you are just making things up.
The correct property is that if a ≤ b, a + c ≤ b + c, and if a ≥ 0 and b ≥ 0, then ab ≥ 0. It is fairly easy to see that these properties hold for dual numbers.
In any case, I'm dropping out of this thread.
Beyond that I'm not sure what to tell you, other than it's fairly easy to see that the dual numbers do satisfy the axioms of an ordered ring that I gave. Here's a large survey of various infinitesimal systems by Philip Ehrlich where he also notes the dual numbers are an ordered ring: https://arxiv.org/pdf/1808.03345.pdf.