The transformation equations you specify are not correct since they do not respect unitarity. The condition of unitarity (or energy conservation) for the action of the beam-splitter gives the following transformations:
$\hat{c}=\sqrt{\tau}\hat{a}+\sqrt{1-\tau}\hat{b}$
$\hat{d}=\sqrt{1-\tau}\hat{a}-\sqrt{\tau}\hat{b}$
The minus sign in the second equation ensures that unitarity is respected.
For reasons that will become clear soon, let us invert these equations to get the input mode operators $\hat{a}$ and $\hat{b}$ in terms of the output mode operators $\hat{c}$ and $\hat{d}$. As expected from arguments of reversibility, we get:
$\hat{a}=\sqrt{\tau}\hat{c}+\sqrt{1-\tau}\hat{d}$
$\hat{b}=\sqrt{1-\tau}\hat{c}-\sqrt{\tau}\hat{d}$
It is useful to look at this problem in the Heisenberg picture where the action of the beam-splitter is entirely on the mode creation and annihilation operators with initial field state being assumed as vacuum.
Since the input states being considered are the Fock states $|m\rangle_{a}$ and $|n\rangle_{b}$ the full intial field state can alternatively be written as:
${(a^{\dagger})^m(b^{\dagger})^n |0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d}}$
Now we substitute the earlier expressions for $\hat{a}$ and $\hat{b}$ in terms of $\hat{c}$ and $\hat{d}$ given by the beam-splitter transformations. The field state after the mode transformations is,
$(\sqrt{\tau}\hat{c}^{\dagger}+\sqrt{1-\tau}\hat{d}^{\dagger})^m(\sqrt{1-\tau}\hat{c}^{\dagger}-\sqrt{\tau}\hat{d}^{\dagger})^n|0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d}$
Thus the output states for a beam-splitter transformation on input Fock states have been obtained.
As Peter Shor correctly pointed out, a beautiful consequence of these transformations is the Hong-Ou-Mandel effect. It states that when single-photon states are incident at the same time on the input ports of the beam-splitter, both photons emerge from the same output port.
This may be verified easily from the equation we have obtained by putting $m=n=1$. Also for convenience let us put $\tau=0.5$ i.e the beam-splitter is $50:50$ ratio. The output field state is,
$\frac{1}{\sqrt{2}}(\hat{c}^{\dagger}+\hat{d}^{\dagger})\frac{1}{\sqrt{2}}(\hat{c}^{\dagger}-\hat{d}^{\dagger})|0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d}$
$=\frac{1}{2}\big((\hat{c}^{\dagger})^2-(\hat{d}^{\dagger})^2\big)|0\rangle_{a}|0\rangle_{b}|0\rangle_{c}|0\rangle_{d}$
$=\frac{1}{\sqrt{2}}(|2\rangle_{c}|0\rangle_{d}-|0\rangle_{c}|2\rangle_{d})$
Thus, we clearly see that either both photons emerge from port $C$ or both emerge from port $D$. Such a state is referred to as a two-photon NOON state (the state looks like that when N=2) and this effect is of paramount importance in linear optical quantum computing schemes.
What does it mean for a particular mode of particles in an infinite square well to have a Fock state $|0\rangle$ with Gaussian wave function?
As far as the many-particle setting is concerned, I am tempted to say that the short answer is "Nothing, really, because the $|0\rangle$ state is not Gaussian." :D
Longer answer: The formal vacuum state of second quantization has nothing to do with the ground state of a harmonic mode in this case. It is rather an abstract state enabling the isomorphism between the (anti)symmetric subspace of the actual N particle Hilbert space and an abstract Hilbert space constructed as a direct product of "particle" or "mode" spaces, each equipped with an individual ladder operator algebra.
One fast way to argue that no vacuum Gaussian wavefunction factors into the construct comes close to validating @MarkMitchison's comment: just check the regular localization probabilities, or even the space correlation functions, on the vacuum state. Say the single-particle eigenfunctions are $\phi_n(x)$ (including the actual ground state!) and the corresponding ladder operators are ${\hat a}_n$, ${\hat a}^\dagger_n$, such that the field operators read
$$
{\hat \psi}^\dagger(x) = \sum_n{\phi^*_n(x){\hat a}^\dagger_n}, \;\;\;{\hat \psi}(x) = \sum_n{\phi_n(x){\hat a}_n}
$$
This gives immediately a null single-particle localization probability on the vacuum:
$$
\rho_0(x) = \langle 0 | {\hat \psi}^\dagger(x) {\hat \psi}(x) |0 \rangle = \sum_{m,n}{\phi^*_m(x)\phi_n(x)\langle 0| {\hat a}^\dagger_m {\hat a}_n| 0\rangle } = 0
$$
In other words, the 2nd quantized vacuum is really … empty. In fact it is empty not only at the single-particle level, but at any k-particle level. Which means that the "probability of measuring a nonzero displacement in the particle's quantum field", as you asked, is really null. Or else, the theory does not provide the means to test the wavefunction of the vacuum state.
Note: one might think of a displaced or squeezed vacuum as a counterexample, but at a closer look nothing much changes because the situation simply transfers to unitarily transformed states and operators/observables.
Best Answer
There are many things to be clarified here.
First, you probably mean "tensor products". If you do, then it's important to distinguish "tensor product of states" and "tensor product of spaces". The Fock space is an infinite-dimensional (or at least multi-dimensional) harmonic oscillator and it is isomorphic to a tensor product of one-dimensional Fock spaces or harmonic oscillators.
However, the tensor product of states is not quite the same thing in the sense that the tensor product space is not composed of purely states that are tensor product states, products of elements from the original spaces. Instead, the tensor product space contains all conceivable linear combinations of the tensor products of the original states.
Now, $|2,0\rangle$ in which two particles are in the same mode is in no way equivalent to a tensor product of $|1,0\rangle$ with itself. By computing the tensor product, we are "extending the number of modes", so a tensor product $|1,0\rangle \otimes |1,0\rangle$ is always something of the form $|1,0,1,0\rangle$. In this 4-mode Hilbert space, the first and third mode may have the same properties, there may even be a symmetry between them, but they are not the same mode.
Concerning the last two questions, there is nothing that could be meaningfully called $$ \frac{1}{\sqrt{2}} (|1a,1b\rangle + |1b,1a\rangle) $$ whether or not you replace the commas by $\rangle |$ – which clearly shouldn't matter (it's just a different typographical convention). This expression of yours uses some completely inconsistent notation. You must first decide what properties $x,y$ of the state the symbol $|x,y\rangle$ denotes and you can't ever permute them. You failed to do so because $x$ sometimes refers to "the first photon" and sometimes to "the second photon", and so on. It just makes no sense.
You could choose an alternative multi-body notation in which the wave function may fail to be symmetric or antisymmetric. If you did so, $x$ would always correspond to the first photon and $y$ would always correspond to the second photon. In this notation, the state $|2,0\rangle$ in the occupation number basis would be written as $|\alpha,\alpha\rangle$ where $\alpha,\beta$ are the modes that had the occupation numbers $2,0$, respectively. No nontrivial superposition has to be constructed in this case to symmetrize the state because the state is already symmetric: the wave functions of both photons are $\alpha$, they are the same.