We can also prove that Yetter-Drinfeld condition implies that $\Psi$ is a braiding as follows.

Let $u, v, w \in V $. Then

\begin{align}
& \Psi_{12} \Psi_{23} \Psi_{12} (u \otimes v \otimes w) \\
& =  \Psi_{12} \Psi_{23}(  (u_{(-1)}.v) \otimes u_{(0)} \otimes w ) \\
& =  \Psi_{12}  (  (u_{(-1)}.v) \otimes (u_{(0)})_{(-1)}.w \otimes (u_{(0)})_{(0)} ) \\
& =  ( u_{(-1)}.v )_{(-1)}.( ( (u_{(0)})_{(-1)} ).w ) \otimes (u_{(-1)}.v)_{(0)} \otimes ( u_{(0)} )_{(0)}  \\
& =  ( (u_{(-1)})_{(1)}.v )_{(-1)}.(  (u_{(-1)})_{(2)}.w ) \otimes ((u_{(-1)})_{(1)}.v)_{(0)} \otimes u_{(0)} \\
& =  ( ((u_{(-1)})_{(1)}.v )_{(-1)} (u_{(-1)})_{(2)} ).w \otimes ((u_{(-1)})_{(1)}.v)_{(0)} \otimes u_{(0)}, 
\end{align}
\begin{align}
& \Psi_{23} \Psi_{12} \Psi_{23} (u \otimes v \otimes w) \\
& = \Psi_{23} \Psi_{12} (u \otimes v_{(-1)}.w \otimes v_{(0)}) \\
& = \Psi_{23} ((u_{(-1)}v_{(-1)}).w \otimes u_{(0)} \otimes v_{(0)}) \\
& = (u_{(-1)} v_{(-1)}).w \otimes (u_{(0)})_{(-1)}.v_{(0)} \otimes ( u_{(0)} )_{(0)} \\
& = ( (u_{(-1)})_{(1)} v_{(-1)}).w \otimes (u_{(-1)})_{(2)}.v_{(0)} \otimes u_{(0)}.      
\end{align}

Therefore the Yetter-Drinfeld condition 
\begin{align}
& (h_{(1)}.v )_{(-1)}   h_{(2)} \otimes (h_{(1)}.v)_{(0)}\\
& = h_{(1)} v_{(-1)} \otimes h_{(2)}.v_{(0)}.
\end{align}
implies that (every $u_{(-1)}$ is some $h \in H$)
\begin{align}
& ((u_{(-1)})_{(1)}.v )_{(-1)} (u_{(-1)})_{(2)} \otimes ((u_{(-1)})_{(1)}.v)_{(0)} \otimes u_{(0)} \\
& = (u_{(-1)})_{(1)} v_{(-1)} \otimes (u_{(-1)})_{(2)}.v_{(0)} \otimes u_{(0)}.
\end{align}
Therefore
\begin{align}
& \Psi_{12} \Psi_{23} \Psi_{12} (u \otimes v \otimes w) = \Psi_{23} \Psi_{12} \Psi_{23} (u \otimes v \otimes w).
\end{align}