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 **there is no**: 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 $R^{op}=R^{-1}$).