I’m not quite sure if I fully understand your question, but I'll try to answer as much as possible.

It is true that $D_\mu$ should always act on some fields, but we should also note that this derivative is really flexible because it depends on the object it acts. If it acts on a field with non-abelian gauge symmetry, then it should be $\partial_\mu - igA_\mu$. If it acts on an operator $F$ living in the $SU(N)$ space and transforming as $UFU^{-1}$ and the covariant derivative should be $\partial_\mu - ig[A_\mu, F]$. And if it acts on a gauge scalar that is invariant under gauge transformation, the covariant derivative is just the ordinary derivative.

However, it should noted that $A_\mu$ itself is not a covariant operator, since it doesn't transform as $UA_\mu U^{-1}$, but with an extra term.

As we have seen, the concept of covariant itself also depends on the objects. Covariant fields and covariant operators transform differently as you have mentioned. From this perspective, $F_{\mu\nu}$ is just an operator, and that's why we need to take the trace of it when we construct a pure gauge field action. The reason why operators transform differently is because if we want $F\Phi$ be covariant, then we require $F'\Phi'=UF\Phi=UFU^{-1}U\Phi$ which shows that if $\Phi$ transform as $U\Phi$, then $F$ must transform as $UFU^{-1}$ so that the net result transform again as a covariant field.

Now, let me say something about $F_{\mu\nu}$. By definition, it is $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu -g[A_\mu, A_\nu]$. Notice that this is not equal to $D_\mu A_\nu - D_\nu A_\mu$ because $A_\mu$ is neither a covariant field nor a covariant operator. You cannot replace $D_\mu$ by $\partial_\mu - igA_\mu$ because it makes no sense to define covariant derivative for an object that is not covariant at all. And the transformation rule of $F_{\mu\nu}$ comes solely from the transformation rule of $A_\mu$. There's nothing related to $D_\mu$ here. And to treat it as an independent gauge field, which is obviously different from field $\Phi$ seeing from perspective of the gauge transformation rules. And you can make action terms from gauge fields alone without acting on any $\Phi$, as long as the end result is a gauge scalar (for example, by taking a trace).

The core concept for gauge theory lies not on the redefinition of the derivative, but on how to preserve a gauge symmetry by combining two types of fields, one for what we normally perceived as real particles and one for interaction carriers. Hence, all other definitions should serve to this goal.