Canonical connection on $\mathcal{A}\times X$

Let $$E\rightarrow X$$ be a vector bundle and let $$\mathcal{A}$$ denote the space of connections on $$E$$. Pulling back $$E$$ by the second projection we obtain a vector bundle $$\mathbb{E}=p_2^*E\rightarrow X$$ over $$\mathcal{A}\times X$$.

There exists a canonical connection $$\mathbb{A}$$ on $$\mathbb{E}$$ which is flat in the $$\mathcal{A}$$ direction and equal to $$A$$ on the slice $$\{A\}\times X$$. We obtain the following curvature: $$\begin{cases} \mathbb{A}^2(v,w)= R_A(v,w) & \text{for v,w\in T_xX} \\ \mathbb{A}^2(\alpha,v)=\alpha(v) & \text{for \alpha\in T_A\mathcal{A},v\in T_xX } \\ \mathbb{A}^2(\alpha,\beta)=0 & \text{for \alpha,\beta \in T_A\mathcal{A} }. \end{cases}$$ These identities can be found in Donaldson's "INFINITE DETERMINANTS, STABLE BUNDLES AND CURVATURE" p236 and Itoh and Nakajima's "Yang-Mills Connections and Einstein-Hermitian Metrics" p451.

I do not understand the pairing of the middle line. How does it follow from the definition?

A vector field $$v$$ on $$X$$ and a vector field $$\alpha$$ on $$\mathcal A$$ give rise to two commuting vector fields on $$\mathcal A\times X$$, denoted by the same letters. Then we may regard $$\alpha$$ as (the pullback of) an element of $$\Omega^1(X;\operatorname{End}E)$$ as well. Write the connection as a covariant derivative $$\nabla$$, choose a section $$s$$ of $$E\to X$$ and compute for its pullback (still denoted $$s$$) that $$\underbrace{\nabla_\alpha\nabla_v}_{=\alpha(v)\in\operatorname{End}E}s-\nabla_v\underbrace{\nabla_\alpha s}_{=0} -\nabla_{\underbrace{[\alpha,v]}_{=0}}s=\alpha(v)s\;.$$ Here we have used the interpretation of $$\alpha$$ as a variation of $$\nabla$$, and we have used that $$\nabla$$ is trivial in the $$\mathcal A$$-direction.