Spin, Pauli, and Heisenberg Please forgive my possible misunderstandings, but I'm a YouTube physics student, and long since forgot the physics I learned in 1982...
I make a few assumptions here...

*

*The Pauli Exclusion principle mandates that, in the first electron shell, one electron must be spin up and one spin down.


*Heisenberg states that an electron is "undefined" as to properties until measured.
If I were to be able to measure one electron, and ascertain it's spin, would that not define the other electron's spin? And if so, would that not mean that the other, unmeasured, electron would have it's probability waveform collapse without having been measured, at least to regards spin?
 A: 

*

*The Pauli Exclusion principle mandates that, in the first electron shell, one electron must be spin up and one spin down.


Yes. It is the results from the anti-symmetry requirement for wave-functions that describe more than one electron (or more than one kind of other fermion).
For example, if your "shell" is an orbital like $\phi(\vec r)$ then the two-electron wave function might look like:
$$
\Psi(\vec r_1, s_1; \vec r_2, s_2) = \phi(\vec r_1)\phi(\vec r_2)\frac{1}{\sqrt{2}}\left(|\uparrow, \downarrow\rangle - |\downarrow,\uparrow\rangle\right)\;.
$$
(Note how the overall wave function is anti-symmetric under particle interchange, but not under interchange of only the spatial parts.)



*Heisenberg states that an electron is "undefined" as to properties until measured.


I don't think he ever stated that "an electron is undefined..." But if so, please provide a citation.
Heisenberg (working in 1925) developed matrix mechanics, wherein he was able to calculate nice things like the emission rate for a charged quantum harmonic oscillator. He wasn't concerned with electron orbits and all that, but rather with experimentally measurable things like emission rates.

If I were to be able to measure one electron, and ascertain it's spin, would that not define the other electron's spin?

Anyways, let's assume you had a way to measure the spin of "electron #1" (despite electrons being indistinguishable, e.g., maybe we actually constructed a spin-0 state of distinguishable particles...). And let's say that you measured the spin to be "up." Given the above example wave-function we could have predicted the probability of this as 50% (Born rule) and given the usual old interpretation of "collapse" the wave function would now be:
$$
\Psi(\vec r_1, s_1; \vec r_2, s_2) = \phi(\vec r_1)\phi(\vec r_2)|\uparrow, \downarrow\rangle\;,
$$
but, I'm not sure that this really tell us much, since we already knew that the state always had one electron's spin "in the opposite direction" as the other. Anyways, these electrons are confined to an atom and you won't know which electron you actually measured to begin with.

And if so, would that not mean that the other, unmeasured, electron would have it's probability waveform collapse without having been measured, at least to regards spin?

The spin part would "collapse" per the usual non-deterministic mechanism of measurements on pure states. But nevertheless, I assure you, you can not use this experiment to instantaneously send information from one place to another (or even at any speed higher the $c$). Sorry.
A: 
If I were to be able to measure one electron, and ascertain it's spin, would that not define the other electron's spin?

Yes.

And if so, would that not mean that the other, unmeasured, electron would have it's probability waveform collapse without having been measured, at least to regards spin?

In a sense, because of the quantum mechanical mathematical correlations, if there are two orbital electrons with opposite spins and you manage to measure one of them, you are actually measuring both.
The measurement destroys the initial wavefunction and a new wavefunction will have to describe the new system. To measure a bound electron's spin you have to interact with it and extract it from the atom.
Once you know the extracted electron spin the spin of the remaining electron in the initial atom is mathematically known. A new wavefunction will describe the ion.
A: 

*

*The Pauli Exclusion principle mandates that, in the first electron shell, one electron must be spin up and one spin down.


This is not exactly right, since there's nothing special about the up-down axis. After all, why shouldn't it instead be the case that one electron has spin left and the other spin right? Instead, we say that the pair of electrons have opposite spins, which can be represented by the state $\frac1{\sqrt{2}}(\lvert\uparrow\downarrow\rangle-\lvert\downarrow\uparrow\rangle)$, common referred to as the "singlet" state. This means that if you measure the two spins along the same axis, you will get opposite results, for any choice of axis.



*Heisenberg states that an electron is "undefined" as to properties until measured.


Based on the tags, I assume that you are referring to the Heisenberg uncertainty principle here, which relates the products of the variances (uncertainties) of two observables to their commutator. While it is commonly interpreted to suggest that the value of an observable is undefined when the system is not in an eigenstate of that observable, there are other ways of interpreting this. You ask

If I were to be able to measure one electron, and ascertain it's spin, would that not define the other electron's spin?

In this question, you miss the important detail that you can only measure the spin along a particular axis, since the components of the spin operator do not commute with each other. Now, we know that if you measure the two spins along the same axis, you will get opposite results, for any choice of axis. Given this, it seems natural to suggest that if you measure the spin of one electron along e.g. the $z$-axis, then you instantly know the spin of the other electron along the $z$-axis. This is where things get interesting. Assuming that this is the case, one might ask, as you did

And if so, would that not mean that the other, unmeasured, electron would have it's probability waveform collapse without having been measured, at least to regards spin?

To some, this would be a reasonable conclusion, while to others, like Einstein, it would not. The reason that Einstein would not have considered it reasonable was because he did not like the idea that measuring one electron would somehow affect the other one, through some sort of "spooky action at a distance." This was because he believe that locality was such an important concept in physics that he did not even consider the possibility of nonlocality. That being said, later on, Bell proved that quantum mechanics is fundamentally nonlocal, so Einstein was wrong in his assumption. More to the point, since the pair of electrons is in an entangled state, we cannot talk about the states of the two electrons separately, so it makes sense that measuring one effectively changes the state of the other, although it's really just transitioning from an entangled state to a separable state.
