Why does ultrasonic frequency sound have more attenuation than low-frequency sound?

In many acoustic power-transfer applications, ultrasonic sound is used for transmission of power. However, the efficiency of transmission over distance reduces as attenuation is high. Why does sound at ultrasonic frequency have high attenuation?

When sound travels through a medium, its intensity diminishes with distance. In idealized materials, sound pressure (signal amplitude) is only reduced by the spreading of the wave. Natural materials, however, all produce an effect which further weakens the sound. This further weakening results from scattering and absorption. Scattering is the reflection of the sound in directions other than its original direction of propagation. Absorption is the conversion of the sound energy to other forms of energy. The combined effect of scattering and absorption is called attenuation. Ultrasonic attenuation is the decay rate of the wave as it propagates through material.

Absorption is a function of frequency--more cycles per second means the material is being 'worked' more--expending more energy into th material. Scattering is also a function of frequency--higher frequency pulses are more more likely to be deflected off in another direction ('specular reflection'--google it).

As user45664 notes, attenuation is the summation of all kinds of losses, including absorption, scattering, or reflection.

Ultrasonic absorption coefficients have been determined over a wide frequency range in all cases and found to be largely independent of the degree of cellular and subcellular destruction, suggesting that more than two-thirds of liver tissue absorption originates on a macro-molecular level including submicroscopic particles such as ribosomes and aggregations of macromolecules.

Let's consider this intuitively for a for a moment. In human tissues there's (typically) a linear increase in the attenuation of ultrasound with frequency from $0.3$ to $10$ MHz. If we take the speed of sound in human soft tissue to be $1540 \frac{m}{s}$, we can estimate the wavelength of the ultrasound:

\begin{align} 3 \cdot10^5 \ \text{Hz} ≤ \ & f \ ≤ 1 \cdot 10^7 \ \text{Hz} \\[10pt] \frac{1540 \frac{m}{s}}{3 \cdot10^5 \ \text{Hz}} ≤ \ & \lambda \ ≤ \frac{1540 \frac{m}{s}}{1 \cdot10^7 \ \text{Hz}} \\[10pt] 5.1 \cdot 10^{-3} \ m ≤ \ & \lambda \ ≤ \ 2.0 \cdot 10^{-4} \ m \\ \end{align}

Which - at the smallest - is still an order of magnitude larger than most eukaryotic cells ($2 \cdot 10^{-5} \ m$), but small enough where it should be apparent that the size of the wave is approximately the same as structures within the tissue. This will tend to cause resonance of those structures, and a loss of acoustic energy proportional to the number of structures along the path.

Waves with lower frequencies will not have as pronounced of an effect.

In addition, frequency matters for scattering. These UNC lecture notes on the Acoustics of the Body are helpful in visualizing why scattering tends to happen more for high-frequencies than low:

For the so called nonspecular boundaries of different tissues (of which the kidney's many nephrons are a great example,) the size of the bumps and crenelations of the tissue boundary scatter wavelengths smaller than them. This also causes increased attenuation for smaller wavelengths.