Let's derive these transformation laws from the first principles.
Start with the gauge transformation.
Let's have a field $\psi(x)$ that transforms in some representation $\rho$ of the gauge group. Then we have,
$$ \psi(x) \mapsto \psi'(x) = \rho(g(x)) \psi(x), $$
parameterized by the group-valued map $g(x)$.
There's a mapping between (at least the neighborhood of the identity element in) the group and the algebra, it's called the exponential map.
Probably the best way to see it is to consider the universal enveloping algebra (UEA) which contains both the Lie algebra and the Lie group.
Working in the UEA, we can write an equation
$$ g(x) = 1 + \varepsilon \alpha(x) + \mathcal{O}(\varepsilon^2), $$
which relates the finite gauge transformation parameterized by $g(x)$ and the infinitesimal transformation parameterised by $\alpha(x)$ and the infinitesimal parameter $\varepsilon$.
The infinitesimal transformation law for $\psi(x)$ is then,
$$ \psi(x) \mapsto \left(1 + \varepsilon \rho(\alpha(x)) \right) \psi(x), $$
$$ \frac{\delta \psi(x)}{\delta \varepsilon} = \rho(\alpha(x)) \psi(x). $$
Now note that the gradient of $\psi$ doesn't transform linearly:
$$ d \psi \mapsto \rho(g) d \psi + d \rho(g) \cdot \psi. $$
To compensate, we introduce the covariant derivative $D \psi = d \psi + \rho(A) \psi$. By construction,
$$ D \psi \mapsto \rho(g) D \psi. $$
As an exercise I will leave deriving the transformation law for $A$ from here.
You already posted it in your comments, so I am sure you're familiar with it:
$$ A \mapsto g A g^{-1} + g dg^{-1}. $$
Note that we can write an equation without $\rho$ here, because the above reasoning is valid for any representation $\rho$. This equation is valid in the UEA, so it's not a problem that we have Lie group and algebra elements in the same equation.
Now to derive the infinitesimal form:
$$ g(x) = 1 + \varepsilon \alpha(x) + \mathcal{O}(\varepsilon^2), $$
$$ A \mapsto \left( 1 + \varepsilon \alpha \right) A \left( 1 - \varepsilon \alpha \right) + \left( 1 + \varepsilon \alpha \right) d \left( - \varepsilon \alpha \right) + \mathcal{O}(\varepsilon^2) = $$
$$ A - \varepsilon [A, \alpha] - \varepsilon d \alpha + \mathcal{O}(\varepsilon^2). $$
$$ \frac{\delta A}{\delta \varepsilon} = - d \alpha - [A, \alpha] = - d \alpha - \rho_{\text{adjoint}}(A) \alpha = - D_A \alpha. $$
Here $D_A \alpha$ is the covariant derivative taken in the adjoint representation.
That's your formula (30) from your post.