Let $F$ be a vector-valued form then $F = (F_1,F_2, \dots , F_n)$ where $F_1,F_2, \dots , F_n$ are functions. The definition of $dF$ is that we exterior differentiate component-wise, $dF = (dF_1,dF_2, \dots , dF_n)$. By definition,
$$ \omega F = (\omega F_1, \omega F_2, \dots , \omega F_n)$$
$$ d(\omega F) = (d(\omega F_1), d(\omega F_2), \dots , d(\omega F_n)).$$
Next, simply apply the usual product rule for forms on each component,
$$ d(\omega F) = (d\omega \wedge F_1-\omega dF_1, d\omega \wedge F_2-\omega dF_2, \dots , d\omega \wedge F_n-\omega dF_n).$$
Finally, factor out the one-form $\omega$ etc.. to find:
$$ d(\omega F) = d\omega \wedge F - \omega \wedge dF.$$
Some of these $\wedge$'s are at times omitted in texts. I think you can see that the same rule holds for $d(F\omega)$ assuming you define right-multiplication on the vectors of forms.