Why is the damping force proportional to $v$ and not $v^2$?

At low velocity $v$ the flow of the fluid around the object is mostly laminar and the drag force a viscous response, which is proportional to $v$.

But at higher velocity, flow becomes turbulent and inertial forces acting on the flowing fluid have to be taken into account. In those conditions the drag force becomes proportional to the square of $v$.

Damped harmonic oscillations are an extremely broad paradigm, and there are many physically dissimilar examples for which the force behaves in completely different ways as a function of velocity.

• In the standard Amontons-Coulomb model of friction, we have $F\propto v^0 \operatorname{sign}(v)$.

• In the case of viscous drag, we have $F\propto v^1$.

• For high velocities, we typically have, approximately, $F\propto v^2\operatorname{sign}(v)$.

The reason that people like to talk about $F\propto v^1$ is not physics, it's simply that the resulting solutions come out to have a simple analytic form. One way to see why the exponent 1 is mathematically special is that in this case, the equations of motion can be put in the form $x''+ax'+bx=0$ (the homogeneous case, i.e., free oscilations). Then we can take $x=e^{rt}$, where $r$ is a complex number, and the solutions correspond to values of $r$ that are roots of a quadratic.