Dear John,
I tried to follow your computation until the first place where I couldn't understand a step. This comes at:
However, $$ P_{u^2_1}(u^1_1u^1_2) = \langle R, u^2_1 \otimes u^1_1u^1_2 \rangle = \sum_z \langle R,u^2_z \otimes u^1_1 \rangle \langle R, u^z_1 \otimes u^1_2 \rangle, $$
Rather than the RHS, I would expect
$$<(\operatorname{id}\otimes \Delta)(R), u^2_1 \otimes u^1_1 \otimes u^1_2> =<R_{13}R_{12}, u^2_1\otimes u^1_1\otimes u^1_2>
=\sum_z<R,u^2_z\otimes u^1_2><R,u^z_1\otimes u^1_1>
$$
which seems different than what you wrote. It seems you have used the opposite comultiplication in your computations so that where I wrote $R_{13}R_{12}$ above, you instead had $R_{12}R_{13}$. I hope this helps. I am aware that pairing of Hopf algebras sometimes requires matching multiplication of $H$ with opposite co-multiplication of $H^*$. However, you seem to be working from Klymik and Schmudgen's text, which doesn't not use opposite co-product in the definition of dual pairing of Hopf algebras.
I haven't checked the details to see if the above resolves your issue. Perhaps this is still not your source of confusion, but it confused me when I first read it in your post.
Looking again at what you wrote, this means that the two computations you did for $P_c(ab)$ and $P_c(ba)$ are thus switched, so that you are multiplying $P_c(ab)$ by $q$ instead of $P_c(ba)$, as you thought. Multiplying instead of dividing by $q$ gives the discrepancy of $q^2$
thanks, -david
