The time dilation due to motion in a circle, relative to an observer at the centre, is just the usual Lorentz time dilation due to the velocity of the motion. If you're interested, in my answer to Is gravitational time dilation different from other forms of time dilation? I showed how this is derived from the metric.
Anyhow, as you say, the time dilation relative to a second observer also in circular motion will be a complex function of time as the relative velocity of the two observers changes. However you can calculate the time dilation for both observers relative to the centre, and then take the ratio to get the average (over many orbits) time dilation between your two observers. If we represent the velocity of the satellite as $v_s$ and the velocity of the observer on the Earth as $v_e$, then the time dilation of the satellite relative to the surface will be:
$$ t_r = \frac{\gamma_e}{\gamma_s} = \frac{\sqrt{1 - v_s^2/c^2}}{\sqrt{1 - v_e^2/c^2}} \tag{1} $$
If you attempt this calculation you'll find your calculator doesn't have enough significant figures to avoid rounding errors (well, probably, at least mine doesn't) but we can use a binomial expansion to approximate $\gamma$:
$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \approx 1 + \frac{1}{2}\frac{v^2}{c^2} $$
or alternatively:
$$ \frac{1}{\gamma} \approx 1 - \frac{1}{2}\frac{v^2}{c^2} $$
Put these approximations into equation (1) and you get:
$$\begin{align}
t_r &\approx (1 + \frac{1}{2}\frac{v_e^2}{c^2})(1 - \frac{1}{2}\frac{v_s^2}{c^2}) \\
&\approx 1 - \frac{v_s^2}{2c^2}\left(1 - \frac{v_e^2}{v_s^2} \right)
\end{align}$$
So we have a correction factor of $1 - v_e^2/v_s^2$. In your example the velocity of the satellite is 4500m/s and the velocity of the faster moving bit of the Earth's surface (at the equator) is 464m/s, so the correction factor is about 1%. The time dilation of the satellite as observed from the equator will be about 1% less than as observed from centre of the Earth.
No the tangential speed of an observer on the surface of the earth and the satellite relative to each other is not the same.
The tangential velocity of the satellite is much more greater than that of an observer on earth's surface, just as you pointed out in the question.
It's the relative angular velocity that is same in both the cases that's zero, as both cover 2pi radians(One revolutin) in 24hrs.
Also note that both of them are accelerating, so the observer seeing the satellite at rest does not imply 0 relative tangential velocity.
Best Answer
You must assume circular orbits, as stated in another answer. The ground speed will also be variable due to inclination, unless it's equatorial. An extreme example is an east-west orbit versus a west-east orbit. In one case, the ground velocity is added, in the other case, it's subtracted.
But with the query limited to equatorial orbits, we can continue. Find the orbital velocity from the radius. Convert the orbital velocity to angular velocity, and then multiply by the Earth's radius. Last, subtract the Earth's movement from this.
$$ v_{gnd} = \sqrt{ \frac{ GM }{ r} } \frac{ 1 }{ r } R_e - 0.465 \frac{ km }{s} \\ = \left( 7.9 \frac{km}{s} \right) \left( \frac{ R_e }{ r } \right)^{3/2} - 0.465 \frac{ km }{s} $$
You can simply plot this. For length units, I'm using multiples of Earth's radius. So r=1 here is the surface.
For inclined orbits, the above relationship will be combined with some trigonometry, and change over the period of the orbit.