We can also prove that Yetter-Drinfeld condition implies that $\Psi$ is a braiding as follows.
Let $u, v \in V, h \in H$. Then
\begin{align} & \Psi_{12} \Psi_{23} \Psi_{12} (u \otimes v \otimes h) \\ & = \Psi_{12} \Psi_{23}( (u_{(-1)}.v) \otimes u_{(0)} \otimes h ) \\ & = ( u_{(-1)}.v )_{(-1)}( ( (u_{(0)})_{(-1)} ).h ) \otimes (u_{(-1)}.v)_{(0)} \otimes ( u_{(0)} )_{(0)} \\ & = ( (u_{(-1)})_{(1)}.v )_{(-1)}( ( (u_{(-1)})_{(2)} ).h ) \otimes ((u_{(-1)})_{(1)}.v)_{(0)} \otimes u_{(0)} (1) \\ & \Psi_{23} \Psi_{12} \Psi_{23} (u \otimes v \otimes h) \\ & = (u_{(-1)} v_{(-1)}).h \otimes (u_{(0)})_{(-1)} \otimes ( u_{(0)} )_{(0)} \\ & = ( (u_{(-1)})_{(1)} v_{(-1)}).h \otimes (u_{(-1)})_{(2)} \otimes u_{(0)}. (2) \end{align}
Therefore the Yetter-Drinfeld condition implies that (1)=(2).