For a Hopf algebra $H$ with antipode $S$, let $M$ be a left $H$module with the action $h \otimes m \mapsto \rho(h,m)$, and also a left $H$comodule with coaction $\delta \colon m \mapsto m^{(1)} \otimes m^{(0)}$. For $M$ to be a YetterDrinfeld module, it must satisfy the compatibility condition $$ \delta(\rho(h,m)) = h_{(1)}m^{(1)} S(h_{(3)}) \otimes \rho(h_{(2)},m^{(0)})\,. $$ Is there a natural commutative diagram to draw that illustrates this compatibility condition? I've included my attempt below in a CW answer. Also, is there any more reason behind this condition besides "it's the condition we need to be true to get the nice braiding to work out in the YetterDrinfeld category?".

1$\begingroup$ As for the last question: The category of YetterDrinfeld modules of $H$ is equivalent to the category of modules over the Drinfeld double $D(H)$. But I guess this merely shifts your question to why the Drinfeld double is defined the way it is. $\endgroup$– Mathematician 42Feb 3, 2018 at 6:06
2 Answers
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.
The equivalent way of writing the YetterDrinfeld compatibility condition between left action and left coaction described in this post can be easily translated into a commutative diagram. It looks like this $$ (\nabla\otimes\rho)(1\otimes\tau\otimes 1)(\Delta\otimes\delta)=(\nabla^\mathrm{op}\otimes 1)(1\otimes\delta)(1\otimes\rho)(\Delta^\mathrm{op}\otimes 1)\,. $$ Similary equations work for ${}_H\mathcal{YD}^H$ instead of ${}_H^H\mathcal{YD}$, see for example Kassel, Definition IX.5.1, page 220. These equations tell us how $\rho$ and $\delta$ "commute". We need precisely this condition to compute the Drinfeld center of ${}_H\mathrm{Mod}$.