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).