Let $\mathcal H$ be a connection on a vector bundle $E\overset{\pi}{\longrightarrow} M$. For $X \in \mathfrak X(M)$, let $\bar X \in \Gamma \mathcal H$ his horizontal lift, let $k(v):=\text{pr}_2(Vv):TE\longrightarrow E$ the connection map. For $U,V \in \mathfrak X(M)$, $\xi \in E$, we define the curvature: $R(U,V)\xi:=k([\bar U,\bar V]_{\xi})$
Problem: Prove that $R$ is linear in the third argument.
I know that $k$ is linear and is homomorphism between $TE\overset{d\pi}{\longrightarrow} TM$ and $E\overset{\pi}{\longrightarrow} M$: let $A(u,v):=u+v$, $\mu_{\lambda}(u):=\lambda u$ be the operations in the fibers of $E$, then $k(dA(u,v))=k(u)+k(v)$, $k(d\mu_{\lambda}(u))=\lambda k(u)$ for $d\pi(u)=d\pi(v)$. So, if for $X,Y\in \Gamma VE$ we had $d\mu_{\lambda}(X|_{\xi})=X|_{\lambda \xi}$, and $dA(X|_{\xi},X|_{\zeta})=X|_{\xi+\zeta}$, then R were linear.
But we have $d\mu_{\lambda}(X|_{\xi})=\lambda X|_{\lambda \xi}$, $dA(X|_{\xi},X|_{\zeta})=2X|_{\xi+\zeta}$. So I find for example $\lambda k(X_{\xi})=k(d\mu_{\lambda}(X_{\xi})=k(\lambda X_{\lambda \xi})=\lambda k(X_{\lambda \xi})$.
In what am I wrong?