differentiation - Generalizing the covariant derivate for gauge theory

Concrete example

gauging the complex scalar field

$\mathcal{L}=(\partial_\mu \phi)(\partial^\mu \phi^*)+m^2 \phi^*\phi$

$\phi(x) \rightarrow e^{-i\Lambda(x)}\phi(x)$

$A_\mu \rightarrow A_\mu + \frac{1}{q}\partial_\mu \Lambda$

The covariant derivative is then

$D_\mu = \partial_\mu + iq A_\mu$

I am having trouble reconciling this with a more general formula for the covariant derivative in a gauge theory from Chapter 11 of Freedman and Van Proeyen’s supergravity textbook which reads

We were given previously in the text, the formula for a symmetry transformation on the gauge field,

but I am struggling to rectify the covariant derivative expression with this prescription of the symmetry transformation on the gauge field.

For details on the nomenclature of this textbook, please see my previous post, Gauge theory formalism.

The only relation I can make with my concrete $U(1)$ example from above, is that the formula for the symmetry transformation on the gauge field from this textbook matches up if I take the coupling $q=1$, since $\Lambda$ takes the place of the symmetry transformation's parameter $e^A$ in the textbook.

However, the formula for the covariant derivative in the $U(1)$ case IS NOT

$\begin{eqnarray*} D_\mu &=& \partial_\mu - \delta(A_\mu) \\ &=& \partial_\mu - \partial_\mu \Lambda \end{eqnarray*}$

as the formula from the textbook prescribes.