Let me put my two cents in here too, as on MSE, together with Ré and Emily answers.
Directly concerning your question about the significance of the axiom involving $\rho$, consider the following: if $A$ is a bialgebra over the field $\Bbbk$, then the category ${{}_A\mathfrak{M}}$ of left $A$-modules is a monoidal category in such a way that the forgetful functor $\omega\colon {{}_A\mathfrak{M}} \to \mathfrak{M}$ to the category of vector spaces is a monoidal functor. That is to say, $\Bbbk$ is a left $A$-module with action provided by $$A \otimes \Bbbk \to \Bbbk, \qquad a \otimes \lambda \mapsto \varepsilon(a)\lambda,$$ and if $M,N$ are left $A$-modules, then $M \otimes N$ is a left $A$-module with action $$A \otimes (M \otimes N) \to M \otimes N, \qquad a \otimes (m \otimes n) \mapsto a_{(1)}\cdot m \otimes a_{(2)} \cdot n.$$ By the compatibility between the coalgebra and the algebra structure on $A$, $(A,\Delta,\varepsilon)$ is a comonoid in the monoidal category $\left({{}_A\mathfrak{M}},\otimes,\Bbbk\right)$ of left $A$-modules.
In this framework, a left Hopf module over $A$ can be defined as a comodule over the comonoid $(A,\Delta,\varepsilon)$ in the monoidal category $\left({{}_A\mathfrak{M}},\otimes,\Bbbk\right)$ and the axiom satisfied by $\rho$ states exactly that $\rho\colon M \to A \otimes M$ is a morphism of left $A$-modules.
Equivalently, you may check that also the category ${{}^A\mathfrak{M}}$ of left $A$-comodules is a monoidal category with $\otimes$ and $\Bbbk$ and that a left Hopf module over $A$ can be also defined as a left module over the monoid $(A,m,u)$ in the monoidal category $\left({{}^A\mathfrak{M}},\otimes,\Bbbk\right)$. In this case, the condition you are looking at is a condition on the action $\mu$ rather than on $\rho$ and it is telling you that $\mu\colon A \otimes M \to M$ is a morphism of left $A$-comodules.
