Let me try to rephrase everything in more modern terms.
First of all $R\in\mathrm{End}(V\otimes V)$. Then let me denote by $U$ the $\mathbb{C}\langle u^i_j \rangle$-valued matrix with entries being $u^i_j$'s ($U\in{\rm End}(V)\langle u^i_j \rangle$).
Now the two-sided ideal $\mathcal J(R)$ is generated by the entries of the matrix $$ M(U):=U_1U_2R-RU_2U_1. $$ This expression lies in ${\rm End}(V\otimes V)\langle u^i_j \rangle\cong{\rm End}(V)\otimes{\rm End}(V)\otimes \mathbb{C}\langle u^i_j \rangle$. Viewing elements of this space as tensors with $3$ components some people rewrite it as follows: $$ U_{1,3}U_{2,3}R_{1,2}-R_{1,2}U_{2,3}U_{1,3}. $$
REMARK: it seems that Klimyk and Schmudgen chose a similar but different $M(U)$, but it is not very important.
Now we want to define $r:\mathbb{C}\langle u^i_j \rangle\otimes \mathbb{C}\langle u^i_j \rangle\to \mathbb{C}$ It is sufficient to define it on generators, and the best to organize the corresponding coefficients is to give an expression for $$ r(U\otimes U)\in{\rm End}(V)\otimes{\rm End}(V)\cong{\rm End}(V\otimes V). $$ We define naively $r(U\otimes U):=R$.
We then need to check that the elements $r(M(U)\otimes U)$ and $r(U\otimes M(U))$, lying in ${\rm End}(V^{\otimes3})$, vanish. Let me try with the second one.
First part: $$ r(U\otimes U_1U_2R)=r(U\otimes U_1U_2)R_{2,3}=r(U\otimes U_1)r(U\otimes U_2)R_{2,3}=R_{1,2}R_{1,3}R_{2,3} $$
Second part: $$ r(U\otimes RU_2U_1)=R_{2,3}r(U\otimes U_2U_1)=R_{2,3}r(U\otimes U_2)r(U\otimes U_1)=R_{2,3}R_{1,3}R_{1,2} $$
So there is no problem here since we find Yang-Baxter.
Let me try now with the first one.
First part: $$ r(U_1U_2R\otimes U)=r(U_1U_2\otimes U)R_{1,2}=r(U_1\otimes U)r(U_2\otimes U)R_{1,2}=R_{1,3}R_{2,3}R_{1,2} $$
Second part: $$ r(RU_2U_1\otimes U)=R_{1,2}r(U_2U_1\otimes U)=R_{1,2}r(U_2\otimes U)r(U_1\otimes U)=R_{1,2}R_{2,3}R_{1,3} $$
There seems to be a problem here since we find an expression which is not Yang-Baxter: $$ R_{1,3}R_{2,3}R_{1,2}-R_{1,2}R_{2,3}R_{1,3} $$ But applying the flip $\tau_{1,2}$ we get $$ R_{2,3}R_{1,3}R_{2,1}-R_{2,1}R_{1,3}R_{2,3} $$ which is an avatar of Yang-Baxter (as far as $R^{op}=R^{-1}$).