Chemical bonding based on hybridisation model
The usual explanation for $sp$ orbitals being more electronegative than $sp^3$ is the fact that more $s$ character means a shorter average distance from the nucleus, thus experiencing a greater effective nuclear charge. This partially explains the decreasing magnitude of the dipole moment of $\ce{CH3-CH2-Cl}$ to $\ce{CH2=CH-Cl}$ to $\ce{HC#C-Cl}$ and the increasing acidity of ethane to ethylene to acetylene.

I agree with you that the example they give doesn't really demonstrate that, since the orientation of the dipole (as you pointed out) relies on the electronegativity of the nitrogen, not the carbon. Specifically, the stability of the left resonance structure below:

$[\ce{^-N=C=Cl+  &lt;-&gt; N#C-Cl}]$

which clearly shows the orientation of the dipole, even if this resonance structure is only a minor contributor.

As a counterexample, chloroacetylene has a dipole with the (-) end on the chlorine, even though the carbon is still $sp$ hybridized. The corresponding resonance structure with a negative charge on C is much higher in energy than in the nitrogen case and so contributes negligibly to the overall structure, so the inductive electron-withdrawing of Cl dominates over the resonance electron-donating.