Proof . Firstly, we can proof that $A$ is a local operator. In fact, since $A$ is linear, thus we only need to show that if for some section $s\in\Gamma(E)$, which vanish at a neighborhood $U$ of $M$ (i.e., for any $p\in U$, $s_p=0$), then $A(s)$ vanish on $U$ also. Without loss of generality, we can assume that $U$ is a neighborhood such that $E|_U$ is trivial (i.e., $E|_U\simeq U\times R^n$, $n=\mathrm{rank} E$). Take a local basis of sections $s_1,s_2,\ldots,s_n$ of $E|_U$, then
$$s=\sum_{k=1}^nf^ks_k.$$
Since $s_p=0$, thus $f^k(p)=0$ for any $p\in U$, therefore $(As)_p=(A\sum_kf^ks_k)_p=(\sum_kf^kAs_k)_p=0$. This proved that $A$ is a local operator, and it is $C^\infty(M)$-linear, thus it can be defined pointwise.

Now, we define a map $\tilde A\in\Gamma(\mathrm{Hom}(E,F))$ (i.e., for any $p\in M$, $\tilde A_p\in\mathrm{Hom}(E_p,F_q)$) as
$$\begin{align*} \tilde A_p\mathpunct{:}E_p&\to F_q\\ v&\mapsto(\tilde Av)=Av, \end{align*}$$
then $\tilde A_p\in\mathrm{Hom}(E_p,F_q)$. Since for any $s\in\Gamma(E)$ such that $s_p=v$, we have $\tilde A_p(v)=(As)_p$, thus $A\in\Gamma(\mathrm{Hom}(E,F))$.

Proof .

1. From the Lemma, We only need to show that
   $$\begin{align*} R^E(f s)&=\nabla^E((\rd f)\otimes s+f\nabla^E s)\\ &=(\rd^2f)\otimes s+(-1)^{|\rd f|}\rd f\wedge\nabla^E s+\rd f\wedge\nabla^E s+f(\nabla^E)^2 s\\ &=-\rd f\wedge \nabla^E s+\rd f\wedge\nabla^E s+fR^E s\\ &=fR^E s. \end{align*}$$
2. for any $s\in\Omega^\cdot(M;E)$, we have
   $$\begin{align*} R^E(X,Y)s&=(R^Es)(X,Y)=(\nabla^E(\nabla^Es))(X,Y)\\ &=\nabla^E_X((\nabla^Es)(Y))-\nabla^E_X((\nabla^Es)(X))-(\nabla^Es)([X,Y])\\ &=\nabla^E_X\nabla^E_Y s-\nabla^E_X\nabla^E_Y s-\nabla^E_{[X,Y]}s. \end{align*}$$
3. Note that the connection $\nabla^E$ on $E$ can naturally induce a connection on $\mathrm{End}(E)$, which is still denoted by $\nabla^E$, as
   $$\nabla^E A=[\nabla^E,A]=\nabla^EA-A\nabla^E.$$
   Then,
   $$\nabla^E(As)=(\nabla^EA-A\nabla^E)s+A(\nabla^E s)=(\nabla^EA)s+A(\nabla^E s),$$
   from which, the third item is trivial.