Your two equations are equivalent, and are both versions of the quantum YBE. (The question from the comments does a good job of answering your classical versus quantum question.)
Write the first as
$$
(c \otimes id)(id \otimes c)(c \otimes id) = (id \otimes c)(c \otimes id)(c \otimes id)
$$
and the second as
$$
R_{12}R_{13}R_{23} = R_{23}R_{13}R_{12}.
$$
Let $\tau : V \otimes V \to V \otimes V$ be the permutation map $\tau(v_1 \otimes v_2) = v_2 \otimes v_1$.
Then a solution $R$ to the second gives a solution $c = \tau R$ to the first, and vice-versa.
Whether or not to include the permutation in the definition of $R$-matrix is just a convention.

In the quantum groups literature, $V$ is a module for a quasitriangular Hopf algebra $H$ and $R$ can be given by the action of the *universal $R$-matrix* $\mathcal{R} \in H \otimes H$. Since $\tau$ will not be given by the action of something in $H \otimes H$, it's natural to leave it out.
(I think the YBE for $R$ is also more natural in the spin chain context, but I know less about that.)

However, in quantum topology we focus more on $c = \tau R$ (because it gives braid group representations) and less on $R$, so some authors eliminate the distinction and call $c$ an $R$-matrix. I prefer to call $c$ a *braiding* and $R$ an *$R$-matrix*, but this convention is not universal.