Consider a principal bundle $P\rightarrow M$ and the associated vector bundle $P\times_{\rho}V$ over $M$ such that $(p,v)=(pg^{-1},\rho(g)v)$. The connection on the principal bundle $A$ defines a covariant derivative on $P\times_{\rho}V$ as follows.
View a section $s$ on $P\times_{\rho}V$ as a $G$-equivariant map $s^{P}:P\rightarrow V$. This means $S^{P}(pg^{-1})=\rho(g)s^{P}(p)$. This defines a $G$-equivariant homomorphism $s^{P}_{*}$ from $TP$ to $V$.
Let $x\in M$ and $v\in TM_{x}$, $H_{A}$ be the horizontal vector bundle in $TP$ defined by $A$. Then for any $p$ be the inverse image of $x$, there is a unique horizontal vector $v_{A}\in H_{A}|_{p}$ being the horizontal lift of $v$ such that $\pi_{*}v_{A}=v$. The covariant derivative $\nabla_{A}$ sends section $v$ to the equivalence of the pair $(p,s^{P}_{*}v_{A})$. A routine exercise showed this is not dependent on the choice of $p$.
Taubes claim in his book Differential Geometry that locally we can write the covariderive $\nabla_{A}$ on $P\times_{\rho}V$ as $$x\rightarrow (x,ds_{U}+\rho_{*}(a_{U})s_{U})$$where $\rho_{*}:\mathcal{G}\rightarrow End(V)$ is the differential of $\rho$ at the identity. I do not understand how he derived this identity given the above definition of $\nabla_{A}$. Even if we use the canonical identification $$\phi_{U}^{*}H_{A}=(x,-g^{-1}a_{U}(v)g)\in TU\oplus \mathcal{G}$$ where $\phi_{U}^{*}$ is the local trivialization. I still do not know how to derive the desired identity. Note in particular the connection 1-form on $P$ is given by $$g^{-1}dg+g^{-1}a_{U}g$$ where $a_{U}:U\rightarrow \mathcal{G}\otimes T^{*}M$. However I do not know how to put all these formulas together.