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?
 A: 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.
from:
https://www.nde-ed.org/EducationResources/CommunityCollege/Ultrasonics/Physics/attenuation.php
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).
A: As user45664 notes, attenuation is the summation of all kinds of losses, including absorption, scattering, or reflection.  
There's good evidence that ultrasound is absorbed by macro-molecules in human tissues:

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.
