For two time-independent vector fields $F,G:\mathbb{R}^D \to \mathbb{R}^D$, the corresponding flows $\Phi^F$ and $\Phi^G$ commute whenever the Lie-bracket $[F,G]=0$ (assuming $F$ and $G$ are complete vector fields).
Is there a similar statement for time-dependent vector fields $f,g:\mathbb{R}^D \times \mathbb{R} \to \mathbb{R}^D$? Is it enough if $f$ and $g$ commute for all $t$?
What I did: I lifted $f$ and $g$ to $\mathbb{R}^{D+1}$ by introducing an additional time variable $t$ and defining $F=(1,f)$. The same for $G$. Then, $[F,G]=0$ if and only if
$$\frac{\partial g}{\partial t} - \frac{\partial f}{\partial t} + [f,g] =0 $$
where $[f,g]$ is the Lie-bracket of $f$ and $g$ for some fixed $t$. I find it strange/curious that $[f,g]=0$ for all $t$ is not sufficient and we need additionally that $\frac{\partial f}{\partial t} = \frac{\partial g}{\partial t} $. I don't trust my derivations and wonder if someone knows something about the commutation of flows for time-dependent vector fields.