Centrifugal Force in the Navier Stokes Equations
Based on the state of the discussion in the comments I was not sure whether this has an answer now, but I stumbled over it, and I think I can give a short answer from which one can learn a lot.

I'll base my discussion on polar $(\rho,\phi)$ coordinates, for the sake of clarity. In classical mechanics we use for the position vector $$\vec r= r \vec{e}_r = r \; (\cos\phi, \sin\phi),$$ because $\dot{\vec{e}}_r = \dot{\phi} \; (-\sin\phi, \cos\phi) = \dot{\phi} \; \vec{e}_{\phi}$. In 3D spherical coordinates, we need a few more of those identities, but the algebra remains the same. From this follows $$\dot{\vec r}=\dot{r}\vec{e}_r + r\dot{\phi}\vec{e}_{\phi} $$ and finally
$$\ddot{\vec r}=(\ddot{r}-r \dot{\phi}^2)\vec{e}_r + (r\ddot{\phi} + 2\dot{r} \dot{\phi})\vec{e}_{\phi} $$
where we have already taken into account the change of coordinates, as well as unit vectors.
The next step would decompose $\dot\phi$ as $\dot\phi = \Omega + u/r$, whereas $$r\dot\phi^2 = r\Omega^2 + 2\Omega u + u^2/r$$
where we identify (in order of the terms) 1.) the centrifugal term, 2.) the coriolis term and 3.) the "curvature/metric" term. This procedure works wonderfully also in spherical coordinates and one recovers perfectly the terms that i.e. Holton or Vallis get in their books via lengthy and opaque 'proof-by-picture' methods.

I'm saying the latter because now I draw conclusions for the complete equations, from my derivations which was done from the incomplete ones:
Conclusions:


Centrifugal terms are not what meteorologists call 'curvature/metric' terms. They can seem the same, if one is sloppy and writes $\Omega=vr$. Or what is often done is in order to calculate the orbit of a satellite to set centrifugal force and gravity equal, whereas one uses $u^2/r$, which is in fact the 'curvature' not the centrifugal term. Also the physically metric term is already $r\dot{\phi}^2$, as this one arises from having a curved coordinate system.  
As pointed out in the comments, the centrifugal term $r\Omega^2$ is customarily hidden in meteorology with the gravity. We would suspect this happens also in your set, because the Coriolis term appears, which hints that the above computation has been carried out to first order and thus the centrifugal term must float around somewhere.
The curvature terms are nowhere to be found in your set of equations. This is probably due to the terms simply being neglected. Also when viewn in their full form in spherical coordinates, they seem quite messy and I didn't find any convenient way of expressing them through a nice vector-or matrix operation. Therefore I strongly suspect, that people when giving the vectorial form of the NS equations usually neglect them, and only add them when presenting the NS equations in component form.
The above minimalistic derivation makes it clear that the curvature, centrifugal and Coriolis terms originate from having a curved coordinate system. Any curved coordinate system is necessarily non-inertial. Thus in Cartesian coordinates, which are inertial, those terms must necessarily vanish.