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The calculation goes as follows: \begin{align} d_A \circ d_A(\alpha) & = d_A(d\alpha + [A, \alpha]) \\ & = d(d\alpha + [A,\alpha]) + [A, d\alpha + [A,\alpha]] \\ & = d^2 \alpha + d[A, \alpha] + [A, d\alpha] + [A, [A, \alpha]] \\ & = 0 + [dA, \alpha] - [A, d\alpha] + [A, d\alpha] + [[A, A], \alpha] \\ & = [dA + [A,A], \alpha] \\ & = [d_A A, \alpha]. \end{align} There's a couple basic identities you need to check in the process, but it's nothing difficult.

Simply use the definition of the derivative in an affine space (since $\mathcal{A}$ is affine): \begin{align} (T_A F)(\phi) & = \lim_{t \to 0} \frac{1}{t} (F(A + t\phi) - F(A)) \\ & = \lim_{t \to 0} \frac{1}{t}(d_{A + t\phi} (A + t\phi) - d_A A) \\ & = \lim_{t \to 0} \frac{1}{t}(d(A + t\phi) + [A, A + t\phi] - dA - [A,A]) \\ & = \lim_{t \to 0} \frac{1}{t}(dA + td\phi + [A,A] + t[A, \phi] - dA - [A,A]) \\ & = d\phi + [A, \phi] \\ & = d_A \phi. \end{align}