Dipole Moment

$BF_3$ (trigonal planar), $BeF_2$ (linear), $CO_2$ (linear) and 1,4-dichlorobenzene (para, symmetrical) all have bond dipoles that cancel, so $\mu = 0$.
HF, $NF_3$, $H_2O$, $NH_3$ and 1,3-dichlorobenzene are polar.

$BF_3$ is symmetrical trigonal planar ($\mu=0$). In $NH_3$ the lone pair moment adds to the N-H bond moments, while in $NF_3$ it opposes the N-F bond moments, so $NH_3>NF_3$. $H_2O$ has the highest dipole moment. Order: $BF_3<NF_3<NH_3<H_2O$.

$CO_2$ and $CH_4$ are symmetrical, so $\mu = 0$.
In $NH_3$ the bond dipoles add to the lone pair moment ($\mu$ = 1.4 D); in $NF_3$ they oppose it ($\mu$ = 0.23 D).
So $NH_3$ has the maximum dipole moment.

An unsymmetrical distribution of the electron cloud makes a molecule polar.
$XeF_4$ (square planar), $BF_3$ (trigonal planar) and $SiF_4$ (tetrahedral) are symmetrical, $\mu = 0$.
$SF_4$ has a see-saw shape with one lone pair, so $\mu \ne 0$.

Both $NH_3$ and $NF_3$ are pyramidal, with a lone pair of electrons on nitrogen.
The net dipole moment is the vector sum of the bond dipoles and the orbital (atomic) dipole due to the lone pair, which points away from the nitrogen nucleus.
In $NH_3$, nitrogen is more electronegative than hydrogen, so each N-H bond dipole points towards nitrogen. The resultant of the three bond dipoles is in the same direction as the lone pair dipole, so they add up and give a large dipole moment.
In $NF_3$, fluorine is more electronegative than nitrogen, so each N-F bond dipole points towards fluorine, away from nitrogen. The resultant of the bond dipoles is opposite to the lone pair dipole, so they partly cancel and the dipole moment is small.
Hence in $NH_3$ the atomic dipole and bond dipole are in the same direction, whereas in $NF_3$ these are in opposite directions.