I am confused about the expression $$F_{\mu \nu} \to F_{\mu \nu}' = U F_{\mu \nu}U^{\dagger}.$$ I found [this](https://physics.stackexchange.com/q/76751/) answer on StackExchange, but I would appreciate some clarification why that analysis makes sense.

To specify notation:
Let $T^a$ be the generators of the gauge group.
We let $U = \exp{i\theta^a T^a}$, and define $$D_{\mu}\psi = \partial_{\mu}\psi - ig A_{\mu}\psi.$$ By demanding that $$D_{\mu}\psi \to D_{\mu}'\psi' = U D_{\mu}\psi,$$ we can deduce that $$A_{\mu} \to A_{\mu}' = U A_{\mu}U^{\dagger} - \frac{i}{g}(\partial_{\mu}U)U^{\dagger}.$$

However, I don't know exactly what $U F_{\mu \nu}U^{\dagger}$ means. Since the $D_{\mu}$ is a differential operator, shouldn't it be defined by how it acts on something? If not, is the expression not ambiguous? How far to the right do operators act? The linked answer acts as if $D_{\mu}$ is a matrix.