Apparently I figured it out.
ANSWER
Previously on pg. 14 we determined using ladder operators that $\psi(4,2,0,0) = \sqrt{3/14} (2^+, 0^-) + \sqrt{8/14} (1^+,1^-) - \sqrt{3/14}(2^-,0^+)$ and $\psi(3,2,1,0) = sqrt{1/2} [(2^+,0^-) + (2^-,0^+)]$.
For the singlet D term, we have $\psi(2,2,0,0) = a (2^+,0^-) + b(2^-,0^+) + c(1^+,1^-)$.
And if we take $\psi(2,2,0,0) \cdot \psi(4,2,0,0) = \sqrt{3/14}a + \sqrt{8/14}c - \sqrt{3/14}b = 0$ because he has said that these micro states must be orthogonal. Multiply by $\sqrt{14}$ to obtain his result. It also follows that $\psi(2,2,0,0) \cdot \psi(3,2,1,0) = \sqrt{1/2} a + \sqrt{1/2} b = 0$. Multiplication by $\sqrt{2}$ will give his result of $a + b = 0$. Also realize that $a^2 + b^2 + c^2 = 1$ must be true as a normalizing condition. Now we have three equations and three unknowns, so solving it is simple.
$\psi(3,2,1,0)$ and $\psi(4,2,0,0)$ were picked in the first place because we had three unknowns and needed three equations to solve it. These wave functions must be orthogonal in the first place, therefore it was a matter of convenience AND that they were made up of the corresponding micro states in the first place.
There is overlap with other questions linked in the comments. But, perhaps the focus of this question is different enough to merit a separate answer. There are at least two distinct but equivalent formalisms of QFT, the canonical approach and the path integral approach. Although, they are equivalent mathematically and in their experimental predictions, they do provide very different ways of thinking about QFT phenomena. The one most suited for your question is the path integral approach.
In the path integral approach, to describe an experiment we start with the field in one configuration and then we work out the amplitude for the field to evolve to another definite configuration that represents a possible measurement in the experiment. So in the two slit case we can start with a plane wave in front of the two slits representing the experiment starting with an electron of a particular momentum. Then our final configuration will be a delta function at the screen representing the electron measured at that point at some later specified time. We can work out the probability for this to occur by evaluating the amplitude for the field to evolve between the initial and final configuration in all possible ways. We then sum these amplitudes and take the norm in the usual QM way.
So in this approach there are no particles, just excitations in the field.
Best Answer
Somewhat analogously to the basis of position eigenstates in QM, there is a basis of field eigenstates in QFT. Each eigenstate is in correspondence to not just one value, but the configuration of the entire field, meaning all field values at all values of $x$.
I will use the real scalar field $\hat{\phi}(x)$ rather than the vector field $A_\mu$, for a simple example. The eigenstates $|\phi(x) \rangle$ are simultaneous eigenstates of every field operator $\hat{\phi}(x)$, namely for every vector $x$ we have an eigenvalue equation. Using $x=1, 3.14$ as symbolic examples (really they should be 4-vectors, so four numbers): $$\hat{\phi}(1) | \phi(x) \rangle = \phi(1) | \phi(x) \rangle $$ $$\hat{\phi}(3.14) | \phi(x) \rangle = \phi(3.14) | \phi(x) \rangle $$ $$...$$
Here, $\hat{\phi}(x)$ denotes the operator of the scalar field. Without the hat, $\phi(x)$ is the eigenvalue corresponding to the operator at $x$ for the eigenvector $|\phi(x) \rangle $. The notation can be a bit confusing here: There is one eigenvector per field configuration, so for every possible function $\phi(x)$ we have a different eigenvector. But given a single eigenvector, it is a simultaneous eigenstate of infinitely many field operators, one for each $x$.
What role do these field eigenstates play in QFT? Not the same as $x$ in QM. You use them to derive the path integral in the theory. However, I have never seen an application in which they were used to predict predict probabilities of measurement outcomes directly through an inner product.
So, if these field eigenstates are not used to directly predict probabilities in QFT, how are probabilities predicted? Fundamentally they use a postulate from Quantum Mechanics, namely that for a system in state $|\psi \rangle$, the probability of finding $n$ particles with momentum $p_1, p_2, ... p_n$ is
$$|\langle p_1 p_2 ... p_n |\psi \rangle|^2 d^4p_1 d^4p_2 ... d^4 p_n$$
It is from this postulate that the scattering cross section is derived.
Finally, then, what about probabilities for positions? Better not ask, as there isn't a good answer to that.