Does the wavelength of a particle depend on the relative motion of the particle and the observer?
So does this mean that the wavelength of a particle depends upon the relative motion between the particle and the observer?
The second statement is more or less equivalent to the first.
Side note: The strange thing about this relationship is that it means that, e.g., wave train seen by one observer as consisting of one wavelength can be seen by another as containing some other number of wavelengths, say 3. This would be impossible if the wave was a real number whose phase was directly were observable, because you could, for example, count nodes.
Yes, the de Broglie wavelength of a particle depends on the relative velocity between the particle and an observer. I find it easier to think about the de Broglie frequency instead of the wavelength. They are related by $v=f\lambda$ or $f=v/\lambda,$ where $v=p/m.$
If the particle is moving towards the observer, the frequency appears higher, and if the particle is moving away from the observer, the frequency appears lower. This is just the Doppler effect. Therefore, if the particle is moving towards the observer, the wavelength appears shorter, and if the particle is moving away from the observer, the wavelength appears to be longer. As you said, this will have implications for interactions because the energy is related to the wavelength.
Your understanding is correct: the de Broeglie wavelength of a particle as measured by an observer depends on the relative motion of the observer and particle. It doesn't make a lot of sense to state that an interaction depends on the relative wavelengths of two particles, because an observer on either particle will perceive the wavelength of his particle as infinite.
However, an external observer watching the two particles will see each as having its own wavelength, and from the wavelengths (and masses) can deduce their momenta and kinetic energies as measured from any moving frame. The kinetic energy in the rest frame of the center of mass of the two particles will be the same as calculated from any other frame.
Note that the above is not quantum mechanically precise.