Let's take it a step at a time:
From my understanding, every particle of matter has a probability wave which fills the entire universe, that's to say the only reason a particle in my fingernail is in my fingernail is because that's where the highest probability of it being is (with the chance that it's anywhere else in the universe being infinitesimally small),
true
so my question is what is a light wave? Also from my understanding, a light wave is simply a probability wave for the location of a photon, and only upon observation does that wave collapse and does the photon 'take on' a definite position in space,
A light wave is also a particle, called the photon. Depending on the observation it can display its particle nature or its wave nature.
so if I compare a particle of matter to a particle of energy (a photon in this case),
a photon is a particle too, not different than other particles
how come the particle of matter has a probability wave that fills the entire universe, and the photon have only a small defined wave which virtually makes up a straight line in space?
A photon has the same behavior as any other particle, depending on the experiment: it either displays its wave nature, or its particle nature depending on the observation: through two slits shows interference, therefore wave properties, absorbed in an atom, particle. Both particles and photons behave the same.
Is the light wave just an area (straight line?) of space where the probability is magnitudes higher than everywhere else and the photon's probability wave is still permeating the rest of the universe,
a volume , and a very small one, like all particles
or is the photon's location DEFINITELY somewhere in the area of space defined by the light wave?
There is no definite in quantum mechanics. Only after the experiment the values are known. Only probabilities can be calculated for individual particles, photons or not.
For example, could a photon that's part of a laser beam ever interact with a photon from a laser beam travelling parallel to the first beam?
Here we are entering deeper into quantum theory, into what is called "second quantization".
Yes, two photons parallel to each other in their probability paths can interact with tiny tiny probabilities by exchanging quanta of energy between them, these are carried by virtual particles covering the whole spectrum that is permitted by quantum number conservations.
Digressing into lasers:
Lasers are another story, and are the result of the possibility that atomic physics allows to have coherent waves of photons. Coherent means "in step" in time and space. Atomic physics because the transitions between energy states in atoms happen with photon exchanges and interactions, and one can easily make a coherent beam of photons, all in phase with each other, as is the laser beam.
A way for a layman to understand the possibilities of the effects of coherence is the famous "soldiers crossing a bridge". In olden times, bridges were mainly stable because the arches held them up statically. The connecting mortar did not have much structural strength. Soldiers when crossing a bridge had to break step, otherwise the hitting of boots in phase could destroy the bridge by building up coherently large forces that the arch could not hold.
In order for the intensity of a light source to stay the same, while each lower frequency photon carries less energy, there must be a greater number (per time, per area) of the lower frequency photons in the beam than the original number of higher frequency photons.
As for the second part of your question, I admit that it can be confusing that the power transmitted by E&M waves depends on the amplitude of the wave, while the power transmitted by a mode of a vibrating string depends on both the amplitude and the frequency of the wave. Ultimately this comes down to fundamental differences in the physics of each wave phenomenon.
The energy in a vibrating string is reducible to the kinetic energy of the moving string elements and the potential energy from the tension felt by each element due to the position of its neighbors. So, at fixed amplitude, you can see that you get even more energy if you jiggle the rope faster.
The energy in an E&M wave is a different effect entirely: it comes from the average size of the (squared) electric field in the wave that can do work to move charged particles. At a fixed amplitude, if you increase the frequency you won't increase the average size of the field.
Best Answer
The dispersion of a wave is a result of the relationship between its frequency and its wavelength, which is appropriately known as the dispersion relation for the wave. For classical waves this depends on the medium: light, for example, will be dispersionless in vacuum and will have dispersion inside material media because the medium affects the dispersion relation. Quantum mechanical waves, on the other hand, have dispersion fundamentally built in.
Let's have an equation look at how light behaves. The dispersion relation for light is $$\omega=\frac c n k,$$ where $k=2\pi/\lambda$ is the wavenumber, $c$ is the speed of light in vacuum, and $n=\sqrt{\varepsilon_r\mu_r}$ is the medium's refractive index. In vacuum, $n\equiv1$ and there is no dispersion: the phase and group velocities, $\frac \omega k$ and $\frac{d\omega}{dk}$ are equal, constant, and independent of $k$, which are the mathematical conditions for dispersionless waves. In material media, though, $n$ will depend on the wavelength - it has to depend on the wavelength - and there will be dispersion.
Matter waves, on the other hand, are quite different. What are their frequency and wavelength, anyway? Well, the first is given by Planck's postulate that $E=h\nu$, and the second by de Broglie's relation $p=h/\lambda$; both should really be phrased as $$E=\hbar \omega\text{ and }p=\hbar k.$$ How are $\omega$ and $k$ related? The same way that $E$ and $p$ are: for nonrelativistic mechanics, as $E=\frac{p^2}{2m}$. Thus the dispersion relation for matter waves in free space reads $$\hbar\omega=\frac{\hbar^2k^2}{2m},\text{ or }\omega=\frac \hbar{2m}k^2.$$ Note how different this is to the one above! The phase velocity is now $v_\phi=\frac\omega k=\frac{\hbar k}{2m}$, and it is different for different wavelengths.
Now, why exactly does that imply dispersion?
Let's first look at the phase velocity, which is the velocity of the wavefronts, which are the planes that have constant phase. Since the phase goes as $e^{i(kx-\omega t)}$, the phase velocity is $\omega/k$. This, of course, is for a single plane wave, and doesn't apply to a general wavepacket for which phase is not as well defined, and for which the different wavefronts might be doing different speeds.
How then do we deal with wavepackets? The approach that works best with the formalism above is to think of a wavepacket $\psi(x,t)$ as a superposition of different plane waves $e^{i(kx-\omega t)}$, each with its own weight $\tilde\psi(k)$: $$\psi(x,0)=\int_{-\infty}^\infty \frac{dk}{\sqrt{2\pi}} \tilde \psi(k)e^{ikx}.\tag{1}$$ Now, if all the different plane-wave components $\tilde \psi(k)e^{ikx}$ moved at the same speed then their sum would just move at that speed and would not change shape.
(More mathematically: if $v_\phi=\omega/k$ is constant, then $$\psi(x,t)=\int_{-\infty}^\infty \frac{dk}{\sqrt{2\pi}} \tilde \psi(k)e^{i(kx-\omega t)} =\int_{-\infty}^\infty \frac{dk}{\sqrt{2\pi}} \tilde \psi(k)e^{ik(x-v_\phi t)} =\psi(x-v_\phi t,0),$$ so the functional form is preserved.)
For a matter wave in free space, however, the different phase speeds are not the same, and the different plane-wave components move at different speed. It is the interference of all these different components that makes them sum to $\psi(x,0)$ in equation (1), and if you mess with the relative phases you will get a different sum. Thus, with longer waves moving slower and shorter ones going faster, wavepackets with lots of detail encoded in long high-$k$ tails of their Fourier transform will change shape very fast.
In general it is hard to predict what the evolution of a wavepacket will do to it in detail. However, it is very clear that all wavepackets will (eventually) spread, since some components are going faster than others. Since the total probability is conserved, this must mean that the probability density will in general decrease.
If I put a particle with zero net momentum localized in some interval, then the probability of it remaining there will decrease. Note, though, that this is no surprise! The Uncertainty Principle demands that there be uncertainty in the particle's momentum. There is then some chance that the particle was moving to the left or to the right, so who's surprised to eventually find it out of the original interval?