To separate time dilation into a gravitational time dilation and a special relativistic time dilation is a somewhat artificial distinction. To calculate the time dilation you calculate the length of the target object's world line, and this is affected both by the curvature of spacetime and the relative motion. Your end result might be split into separate bits depending on the context, but the calculation cannot be split into a gravity bit and a special relativity bit.
A satellite hovering stationary above the earth and one in a circular orbit obviously have very different worldlines, and it shouldn't surprise you to find their time dilation is different. Note that to calculate the time dilation for the orbiting satellite you can't just start with the time dilation for a stationary satellite and just multiply it by some special relativity derived factor. In both cases you have to do a full GR calculation to calculate the satellites proper time.
So if you're describing a satellite in a stable circular orbit calculate the total time dilation using the equation:
$$ t_0 = t_f\sqrt{1 - \frac{3}{2}\frac{r_0}{r}} $$
or for a satellite hovering use:
$$ t_0 = t_f\sqrt{1 - \frac{r_0}{r}} $$
A few extra interesting points (since we're all nerds here :-)
You've spotted that for the orbiting satellite the time dilation becomes infinite at $r = \tfrac{3}{2}r_0$. This radius is known as the photon sphere because it's the distance at which a light ray will orbit a black hole (though note this is an unstable orbit). Since the time dilation becomes infinite as the velocity approaches $c$, or more precisely as the world line approaches a null geodesic, it shouldn't surprise you that your equation gives an infinite time dilation.
Note also that no massive particle can have a stable circular orbit at $r < 3r_0$, despite what the time dilation formula says. Any massive particle attempting an orbit for $r < 3r_0$ will inevitably be drawn into the black hole unless some assistance, e.g. rocket motors, is used to sustain the orbit.
