What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?

I think the simple answer here is resolution.

Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.

If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by

$$lambda = {c over f} $$

so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....

The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate

$$lambda = 0.001 {rm m} = 1 {rm mm}$$

At 20000 Hz $lambda = 75$ mm

Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 discs (or spheres).

The waves are coming from behind the specimen passing through the circle or the waves are reflected on the circle. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 circle, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.

On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
$$lambda /left( {2d} right) = sin left( alpha right) approx alpha ,$$
where $lambda$ corresponds to the wavelength and $d$ is the size of disc (therefore, your discs cannot be closer than this). With a huge, $180^circ$ imaging device one can resolve a distance of $lambda/2$. It can be seen that the achievable resolution (resolvable smallest distance) is proportional to wavelength.

And as it is known,
$$lambda = c/f,$$
where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.

Estimation

You can make an estimation on the achievable resolution. You'd like to do measurements in a human body. That consists of water mainly, the speed of sound in the water is $1500m/s$. At $20text{KHz}$ the resolvable distance is
$$frac{{1500{text{m}}/{text{s}}}}{{2 cdot 20{text{kHz}}}} = 3.75{text{cm}}.$$

Higher frequency provides higher resolution.