In Cliff Taubes' book "Differential geometry", he explains how to define the trivial connection on the product principal bundle $U \times G,$ where $G$ is a Lie group. Namely, the trivial connection $A_0$ is a Lie(G)-valued 1-form which is defined at the point $(p, g) \in U \times G$ by $g^{-1}dg.$
I am confused about what $dg$ is supposed to mean.
Possible interpretation: we could view $g$ as the identity function on the underlying Lie group. Then $dg$ is just the differential, which takes a tangent vector $X \in T_gG$ to itself.
We view $g^{-1}$ as the differential of the group multiplication map $h \mapsto g^{-1}\cdot h.$ Then $g^{-1} dg$ maps $T_gG \to T_eG$ and hence is indeed a Lie algebra value 1-form.
However, this interpretation seems strange because it treats both $dg$ and $g^{-1}$ as differentials, even though the $d$ is missing from the second symbol.
Is my interpretation wrong? If so, could someone provide a correct interpretation?
Your interpretation is correct - it's a common shorthand in this context to write $g^{-1}$ instead of $D(L_g^{-1}).$ The justification is that when $G$ is a matrix group, the differential of left-multiplication by $g$ is again just left-multiplication by $g$.