A.T. said:
Give an example of how something follows from the fact that you call them "forces".
Of course, it doesn't matter what you call them, but the point is that Newton's laws relate motion of one object to vector quantities produced by other objects:
m \frac{\stackrel{\rightarrow}{dU}}{dt} = \stackrel{\rightarrow}{F}
The left-hand side is a fact about the motion of the object, and the right-hand side is about the external situation affecting that motion. In terms of coordinates:
(\frac{\stackrel{\rightarrow}{dU}}{dt})^i = \frac{dU^i}{dt} + sum over j, k of \Gamma^i_{jk} U^j U^k
where \Gamma^i_{jk} are the so-called "connection coefficients", which are due to using nonconstant basis vectors. So the full equations of motion, in terms of components, are:
m(\frac{dU^i}{dt} + sum over j, k of \Gamma^i_{jk} U^j U^k) = F^i
What the idea of "fictitious forces" amounts to is moving the extra terms from the left side (where they describe motion) to the right side (where they are treated as forces):
m \frac{dU^i}{dt} = F^i + F_{inertial}^i
where
F_{inertial}^i = - m sum over j, k of \Gamma^i_{jk} U^j U^k
What difference does it make whether you group it on the left side, or the right side? Well, for one thing, when it comes to figuring out the reaction forces (Newton's third law), only the F^i term is relevant. There are no reaction forces to F_{inertial}^i. For another, since real forces are
vectors, the components transform in a standard way under a coordinate change: If you change coordinates from x^i to y^b, then
F^b = sum over i of \dfrac{\partial y^b}{\partial x^i} F^i
"Inertial forces"
DON'T transform that way.
So sure, you can group whatever terms together you want, and call them whatever you want to call them, but when it comes to reasoning about the physics, you have to separate out the "real" forces from the "inertial" forces. You're basically doing extra steps that have to be undone later.