[step:Verify compatibility on overlaps by computing $(d_A \omega)_\alpha - \psi_{\alpha\beta} (d_A \omega)_\beta$]We claim that on $U_\alpha \cap U_\beta$,
\begin{align*}
(d_A \omega)_\alpha = \psi_{\alpha\beta}\, (d_A \omega)_\beta.
\end{align*}
Start from $\omega_\alpha = \psi_{\alpha\beta}\, \omega_\beta$ and apply $d$:
\begin{align*}
d\omega_\alpha = d(\psi_{\alpha\beta}\, \omega_\beta) = d\psi_{\alpha\beta} \wedge \omega_\beta + \psi_{\alpha\beta}\, d\omega_\beta,
\end{align*}
using the matrix Leibniz rule (the sign is positive because $\psi_{\alpha\beta}$ is a 0-form).
For the second term, using the transformation law $\theta_\alpha = \psi_{\alpha\beta}\, \theta_\beta\, \psi_{\alpha\beta}^{-1} - d\psi_{\alpha\beta}\, \psi_{\alpha\beta}^{-1}$,
\begin{align*}
\theta_\alpha \wedge \omega_\alpha &= \big(\psi_{\alpha\beta}\, \theta_\beta\, \psi_{\alpha\beta}^{-1} - d\psi_{\alpha\beta}\, \psi_{\alpha\beta}^{-1}\big) \wedge \psi_{\alpha\beta}\, \omega_\beta \\
&= \psi_{\alpha\beta}\, \theta_\beta \wedge \omega_\beta - d\psi_{\alpha\beta} \wedge \omega_\beta,
\end{align*}
using $\psi_{\alpha\beta}^{-1} \psi_{\alpha\beta} = I$ in both factors (note that $\psi_{\alpha\beta}$ is a 0-form, so it commutes through the wedge). Summing,
\begin{align*}
(d_A \omega)_\alpha = d\omega_\alpha + \theta_\alpha \wedge \omega_\alpha = d\psi_{\alpha\beta} \wedge \omega_\beta + \psi_{\alpha\beta}\, d\omega_\beta + \psi_{\alpha\beta}\, \theta_\beta \wedge \omega_\beta - d\psi_{\alpha\beta} \wedge \omega_\beta.
\end{align*}
The $\pm d\psi_{\alpha\beta} \wedge \omega_\beta$ terms cancel, leaving
\begin{align*}
(d_A \omega)_\alpha = \psi_{\alpha\beta}\, d\omega_\beta + \psi_{\alpha\beta}\, \theta_\beta \wedge \omega_\beta = \psi_{\alpha\beta}\, (d\omega_\beta + \theta_\beta \wedge \omega_\beta) = \psi_{\alpha\beta}\, (d_A \omega)_\beta.
\end{align*}[/step]