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.
To expand on Xcheckr's answer:
The full equation for a single-frequency traveling wave is
$$f(x,t) = A \sin(2\pi ft - \frac{2\pi}{\lambda}x).$$
where $f$ is the frequency, $t$ is time, $\lambda$ is the wavelength, $A$ is the amplitude, and $x$ is position. This is often written as
$$f(x,t) = A \sin(\omega t - kx)$$
with $\omega = 2\pi f$ and $k = \frac{2\pi}{\lambda}$. If you look at a single point in space (hold $x$ constant), you see that the signal oscillates up and down in time. If you freeze time, (hold $t$ constant), you see the signal oscillates up and down as you move along it in space. If you pick a point on the wave and follow it as time goes forward (hold $f$ constant and let $t$ increase), you have to move in the positive $x$ direction to keep up with the point on the wave.
This only describes a wave of a single frequency. In general, anything of the form
$$f(x,t) = w(\omega t - kx),$$
where $w$ is any function, describes a traveling wave.
Sinusoids turn up very often because the vibrating sources of the disturbances that give rise to sound waves are often well-described by
$$\frac{\partial^2 s}{\partial t^2} = -a^2 s.$$
In this case, $s$ is the distance from some equilibrium position and $a$ is some constant. This describes the motion of a mass on a spring, which is a good model for guitar strings, speaker cones, drum membranes, saxophone reeds, vocal cords, and on and on. The general solution to that equation is
$$s(t) = A\cos(a t) + B\sin(a t).$$
In this equation, one can see that $a$ is the frequency $\omega$ in the traveling wave equations by setting $x$ to a constant value (since the source isn't moving (unless you want to consider Doppler effects)).
For objects more complicated than a mass on a spring, there are multiple $a$ values, so that object can vibrate at multiple frequencies at the same time (think harmonics on a guitar). Figuring out the contributions of each of these frequencies is the purpose of a Fourier transform.
Best Answer
If you get from a Fourier transform the spectra of the square wave you get many wavelengths (and of course frequencies), and it has a bandwidth. Freq and wavelength is defined wrt to the basis waves, which are sinusoids. See below about other basis waves.
If you use it differently it'll be easy to have people confused, or just say you are wrong. The spectrum of the square wave is the delta function convolved with a sync function. In radio and EM transmission and reception they try to filter out the sidelobes (and any freqs outside what you want), and usually the square wave is rounded by some smoothing function.
In QM it's even more critical because the sinusoids are the eigenfunctions of the Hamiltonian (or some physical and important operator that you use with eigenfunctions as the basis for expanding any arbitrary solution. The eigenvalues or for the Hamiltonian the energy, are then we'll defined. A square wave or a limited wave train will not have a definite energy.
This is also reflected in the uncertainty principle in QM, and it is equivalently true for any wave. If your wave last infinitely long your uncertainty in the arrival time is infinite. Then the energy uncertainty has to be zero. In classical waves you need a long train to measure it's frequency accurately, equivalently to QM.
Now, you can choose other basis waves, or in QM, eigenfunctions of any operator (with some constraints) and expand on the basis of a those. The eigenvalues, or values you could actually measure and get a number, will just not be energy. Similarly for classical waves, you can choose mathematically any basis you want. In some signal processing work people use so called wavelets, and other things.