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.
That isn't what an ordered ring is. Your property of a < b → a² < b² doesn't even hold true in the integers. For instance, let a = -2 and b = -1.
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 my argument a and b are positive and this is true for all ordered rings but not for the dual numbers as you've defined them. Specify the ordering and you will realize h can not be larger nor smaller than 0 because both cases lead to a contradiction.
It isn't true for all ordered rings, and the dual numbers are in fact a counterexample to the claim that it is true.
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.
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.