Let $M$ be a Riemannian manifold and $\nabla$ a covariant derivative on a vector bundle $E$. In Heat Kernels and Dirac Operators the connection Laplacian $\Delta$ is defined by
$$\Delta=-\mathrm{Tr}(\nabla^2):=-\mathrm{Tr}(\nabla^{T^*M\otimes E}\nabla^E)$$
and it is said that
\begin{equation}\tag{1}
(\nabla\nabla s)(X,Y)=\nabla_X\nabla_Ys-\nabla_{\nabla_XY}s.
\end{equation}
I think that this follows from the fact that for $T\in \Gamma(M,T^*M\otimes E)$ we have
\begin{equation}\tag{2}
(\nabla T)(X,Y)=\nabla_X(T(Y))-T(\nabla_XY)
\end{equation}
as $(1)$ follows from $(2)$ by plugging in $T=\nabla s$.
In fact I believe that I have a proof of $(2)$. Nevertheless I would appreciate a confirmation that my reasoning and my proof are correct:
\begin{align}
&(\nabla T)(X,Y)=(\nabla_XT)(Y)=\nabla_X(e^i\otimes T_i)(Y)&\\
&=(e^i\otimes\nabla_X T{}_i)(Y)+((\nabla_X e{}^i)\otimes T_i)(Y)&\\
&=e^i(Y)\nabla_X T_i+(\nabla_X e^i)(Y)T_i&\\
&=e^i(Y)\nabla_X T_i+\mathrm{d}_X(e^i(Y))T_i-e^i(\nabla_X Y)T_i&\\
&=Y^i\nabla_X T_i+\mathrm{d}_X(Y^i)T_i-T(\nabla_X Y)&\\
&=\nabla_X(Y^iT_i)-T(\nabla_X Y)=\nabla_X(T(Y))-T(\nabla_X Y)&
\end{align}
Best Answer
Still only a comment - but continuing the above.
So : $\nabla^{T^*\otimes \cal E}(\omega \otimes s) = (\nabla^{T^*} \omega)\otimes s + \omega \otimes \nabla^{\cal E} s$. Be careful with the evaluation on vector fields: $$\nabla^{T^*\otimes \cal E}(\omega \otimes s) (X,Y) = (\nabla^{T^*}_X \omega)(Y)s + \omega(Y) \nabla_X^{\cal E} s.\tag{*}$$
Addendum
Even though I think you had this earlier (from your eliminated comments), and now have it above - I am adding this as promised:
By the product rule fiat/definition,
$$(\nabla_X \omega) (Y) = X (\omega (Y)) - \omega (\nabla_X Y).$$
Therefore, substituting into (*), one has
$$\nabla^{T^*\otimes \cal E}(\omega \otimes s) (X,Y) = {\Large(}X (\omega (Y)) - \omega (\nabla_X Y){\Large)} s + \omega(Y) \nabla_X^{\cal E} s.$$ Rearranging and regrouping, $$\nabla^{T^*\otimes \cal E}(\omega \otimes s) (X,Y) = {\Large(}X (\omega (Y))s + \omega(Y) \nabla_X^{\cal E} s {\Large)} - \omega (\nabla_X Y)s .$$
Therefore $$\nabla^{T^*\otimes \cal E}(\omega \otimes s) (X,Y) = \nabla_X^{\cal E} (\omega (Y) s) - \omega (\nabla_X Y)s, $$
obtaining your desired expression for an elementary / decomposable tensor $\omega \otimes s$ .