Dipole moment in NF3 versus NH3
You should probably be cautious in general applying the concept of bond dipoles to predict molecular dipole moments, and it does seem ad hoc if not downright flimflam to invoke a dipole to the nitrogen lone pair to predict dipole moments in the examples you present (particularly in the case of nitrogen trifluoride). 

Now if you read the wikipedia page on nitrogen trifluoride, it explains that 


  $\ce{NF3}$ is slightly soluble in water without undergoing chemical reaction. It is nonbasic with a low dipole moment of $\pu{0.2340 D}$. By contrast, ammonia is basic and highly polar ($\pu{1.47 D}$).[9] This difference arises from the fluorine atoms acting as electron withdrawing groups, attracting essentially all of the lone pair electrons on the nitrogen atom. $\ce{NF3}$ is a potent yet sluggish oxidizer. 


The low basicity of $\ce{NF3}$ supports the argument that electron density associated with unpaired electrons on N is withdrawn by the fluorine atoms. So it would appear that in the case of $\ce{NF3}$ there might not be a significant dipole moment between the $\ce{N}$ nucleus and an electron pair (or the charge opposite the three fluorines). Instead the electron density (relative to ammonia as a reference) is polarized toward the fluorines. This makes the application of a method of linear summation of bond dipoles somewhat sketchy.

Continuing along similar sketchy lines, the website to which you provided a link argues that the difference in molecular dipoles cannot be explained simply in terms of the order of electronegativity ($\ce{F &gt; N &gt; H}$). But note that electronegativities are computed from dissociation energies, electron affinities, etc, and not from partial charges or dipole moments, and the relation between charges or dipole moments and energies is bound to be highly non-linear, making quantitative prediction of dipole moments based on electronegativities difficult.

What seems necessary is to perform some ab initio electronic structure calculations.