I try to understand the physical definition of covariant derivative in gauge theories in terms of the exterior covariant derivative of vector-valued forms defined as the horizontal projection wrt a connection $\omega$ on a principal $G$-bundle over $\mathbb R^d$, $D^\omega:\Omega^k(P,V)\to\Omega^{k+1}(P,V)$, $D^\omega\eta\equiv(\mathrm d\eta)^H$.
If $\varrho:G\to GL(V)$ is a representation of the (matrix) structure group and $\eta$ is $\varrho$-tensorial, the exterior covariant derivative is given by $$D^{\omega}\eta=\mathrm{d}\eta+\varrho_{*}\omega\wedge\eta. \tag{A}$$ On a (necessarily) trivial bundle over $\mathbb R^d$ we may choose a global section $s$ and consider (A) on the base with $s^*\omega=A$. Any gauge transformation on the trivial bundle may then be identified with a smooth map $g:\mathbb R^d\to G$.
Concretely, I am following Vortices and Monopoles by Jaffe and Taubes. They define the covariant derivative as $D^A\eta=\mathrm d\eta+\varrho_* A\wedge\eta$, where $\varrho=D\varrho_e:\mathfrak g\to\mathfrak{gl}(V)$ is the induced Lie algebra representation. This obviously corresponds to (A).
Now, under a gauge transformation (i.e. a change of section) $g$ the local connection form transforms as $A\mapsto A^g=\mathrm{ad}(g^{-1})A+g^{-1}Dg$ and a the tensorial form $\eta\mapsto\varrho(g)\eta$. They claim that the covariant derivative transform under a gauge transformation as $$D^{A^g}(\varrho(g)\eta)=\varrho(g) D^A\eta.$$ I am not able to verify this for a general representation $\varrho$. What I have tried is to start in local coordinates so that the "$j$-th component" of (A) becomes
$$D^{A^g}_j(\varrho(g)\eta)=\partial_j(\varrho(g)\eta)+\varrho_*(\mathrm{ad}(g^{-1})A_j+g\partial_j g^{-1})(\varrho(g)\eta)\tag{B}$$
If $\eta$ is a section of the adjoint bundle and $\varrho=\mathrm{ad}$ I can use $\mathrm{ad}_*(X)(Y)=[X,Y]$ and explicitly calculate the second term in (A), but in the general case I fail.
Is there an elementary way to explicilty verify the gauge covariance in terms of the components (B)?