The usual explanation for the molecular dipole moment of NF$_3$ being smaller than that of NH$_3$, despite the N-F dipole being stronger than the N-H dipole, is that influence of the lone electron pair on nitrogen is to oppose the net N-F$_3$ dipole, while enhancing that of N-H$_3$. See, for example, this good example.
What bothers me about this explanation is that when the N-F dipole is calculated (experimentally, I think), the effect of the electron pair on N is already factored in. Likewise for the calculation for N-H.
Putting the question in terms of the language of the linked article: "[T]he lone pair's negative concentration will augment the polarity contribution from the polar bonds while for the NF3, the lone pair's dipole would subtract from the dipole contributed by the NF bonds."
If the lone pair on N subtracts from the NF bond in the molecule it must have had the same effect in the original diatomic dipole calculation. Likewise for NH.
If this is true, assuming identical N-F and N-H bond angles in NF$_3$ and NH$_3$, it seems that the vector calculation should still give a stronger dipole moment for NF$_3.$ [Please regard this as a question!]
One thing that would lessen the molecular NF$_3$ dipole moment is a smaller NF bond angle (compared to NH), which increases the lateral component of the dipole at the expense of the axial, making the (axial) resultant smaller. I think fluorine is larger than hydrogen and maybe this would tend to push the fluorine atoms out laterally, decreasing the resultant axial dipole. However I do not see this as an explanation in texts so I am guessing it is wrong.
My confusion is that the vector algebra doesn't seem to support the argument but there is probably something I am not taking into consideration.