Here's my best attempt at a clean commutative diagram. I only got this from decomposing the compatibility condition though, and I don't have any motivation for why this is the diagram that we want to commute.
$$\require{AMScd} \begin{CD} H \otimes M @>{\rho}>> M @>{\delta}>> H \otimes M \\ @V{\Delta^2 \otimes \delta}VV @. @AA{m^2 \otimes \rho}A \\ H^{\otimes 4} \otimes M @>>{\mathbb{1} \otimes T \otimes \mathbb{1}}> H^{\otimes 4} \otimes M @>>{\mathbb{1}\otimes\mathbb{1}\otimes S \otimes\mathbb{1}\otimes\mathbb{1}}> H^{\otimes 4} \otimes M \end{CD} $$
In this diagram, the Hopf algebra is associative and coassociative so the maps $$\Delta^2 \colon H \to H \otimes H \otimes H \quad\text{and}\quad m^2 \colon H \otimes H \otimes H \to H$$ are well defined. The symbol $H^{\otimes 4}$ is short for $H \otimes H \otimes H \otimes H$, and $T$ is an "outer twist" map where $T(a \otimes b \otimes c) = c \otimes b \otimes a$ . Or if you prefer, if $\tau$ is the usual twist map $\tau(a \otimes b) = b \otimes a$, then we can write $T$ as $(\mathbb{1}\otimes{\tau})(\tau\otimes\mathbb{1})(\mathbb{1}\otimes{\tau})$, which I think relates this more closely to the braiding we want in the category.