Kitaev Chain, Interpretation of the states One considers a chain of $N$ sites (open boundary conditions) and corresponding to each site $l$ we have some fermionic annihilation operator $c_l$ and creation operator $c_l^{\dagger}$ (e.g. defined through the Wannier-states for electrons in a solid).
Then one considers the Hamiltonian
$$
H = \sum_{l = 1}^{N-1}\left( c_{l + 1}^{\dagger}c_l + c_l^{\dagger}c_{l+1} + c_l^{\dagger}c_{l + 1}^{\dagger} + c_{l+1}c_l \right)\text{,}
$$
where the constants (giving the dimension $\text{energy})$ are dropped for simplicity. 
Then one introduces the operators
$\eta_l = c_l + c_l^{\dagger}\text{,}\quad \chi_l = \frac{1}{\mathrm{i}}(c_l - c_l^{\dagger})$
for $l = 1, \ldots, N$, satisfying some commutation relations, and another pair of operators
$d_l = \frac{1}{2}(\eta_{l + 1} + \mathrm{i} \chi_l), \quad d_l^{\dagger} = \frac{1}{2}(\eta_{l + 1} - \mathrm{i}\chi_l)$
for $l = 1, \ldots, N-1$, satisfying the canonical anticommutation relations, and then the Hamiltonian $H$ can be rewritten as
$$
H = \sum_{l = 1}^{N - 1} \mathrm{i} \eta_{l + 1} \chi_l = (N - 1) \cdot \text{Id} + 2 \cdot \sum_{l = 1}^{N - 1} d_l^{\dagger}d_l \qquad (\star)\text{.}
$$
Now obtaining the set of equations $(\star)$ is not so difficult, but where I can't follow is the interpretation of this result.
Coming from linear algebra, I would like to interprete this whole procedure as a kind of "basis change". (There are quotation marks, since usually one mixes the basis-elements (corrensponding to the $c_l^{\dagger}$-operators) among each other to form new states. But now we are mixing creation and annihilation operators among each other.
Question 1: Is there any ambiguity when I write $| 0 \rangle \equiv | \text{vac.} \rangle$ due to the transformations we made?
Question 2: If there is no ambiguity, does it hold that $| \text{g.s.} \rangle = | 0 \rangle$ ? (Since applying $d_l$ on $|0\rangle$ by re-expressing it in terms of $c_l$ and $c_l^{\dagger}$ gives zero?)
Question 3: If there is no ambiguity when one writes $| 0 \rangle$, then how can one conclude from $(\star)$ that the ground-state degeneracy is precisely four? 
(In my notes it states that this follows, since $|\text{g.s.}\rangle$, $\chi_N |\text{g.s.} \rangle$, $\eta_1 |\text{g.s.}\rangle$ and $\chi_N \eta_1 |\text{g.s.}\rangle$ all give the same energy, since $[H, \eta_1] = [H, \chi_N] = 0$, but the fact that these operators commute with the Hamiltonian seem to me a little bit accidental. Why can't it be that there are more Operators $O_1, \ldots, O_p$ that also commute with the Hamiltonian, giving a higher degeneracy?)
As a preparation of the fourth question: Let's say corresponding to $c_l^{\dagger}$ we have a Wannier-function $\psi_l(\cdot)$. Then 
$$\langle \, \mathbf{x} \, | \, c_l^{\dagger} \, | \, 0 \, \rangle = \psi_l(\mathbf{x}) \qquad (\star \star) \text{.}
$$
Question 4:
Now what is $\langle \, \mathbf{x} \, | \, d_l^{\dagger} \, | \, 0 \, \rangle = $ ? 
Is it just given by re-expressing $d_l^{\dagger}$ in term of the creation (and annihilation) operators $c_l^{\dagger}$ and then using the definition $(\star \star)$? (Meaning that the answer of question 1 is "There is no ambiguity."?)
 A: The ground state, which serves as the "vacuum" of the theory, is certainly not the empty state with no fermions $|0\rangle$. This is probably the central point to understand here. Kitaev's chain describes a superconductor (specifically, a 1d spinless p-wave superconductor), which means that the ground state consists of Cooper pairs.
I think that the way to approach the problem, conceptually, is the following: we start by artificially doubling the degrees of freedom, treating $c$ and $c^{\dagger}$ as independent variables, and getting a BdG Hamiltonian which we can diagonalize. This is indeed a basis change, as you write, but instead of a basis change of the $N$ original degrees of freedom it is a change of the $2N$ degrees of freedom. So you get $2N$ new creation and annihilation operators, which are related to each other (due to the doubling of degrees of freedom), such that a "creation" operator which corresponds to negative energy is the same (i.e. has the same operatoric content) as a annihilation operator which corresponds to the positive energy (the spectrum will always be symmetric about $E=0$).
Now you can proceed to find the groundstate. It is defined as the state for which all annihilation operators of positive energy map to zero. Which means that we can create it by
$$ |{\rm gs}\rangle = \prod_{E>0}d_E|0\rangle = \prod_{E<0} d^{\dagger}_E|0\rangle$$
We basically filled all Fock states with negative energy. Note that since $d_{E}$ has both $c$ and $c^{\dagger}$ its operation on $|0\rangle$ is not trivial. So there is no ambiguity as to the ground-state (up to the degeneracy). It is just not the empty state.
So far this is true to all superconductors, and not only for Kitaev's chain. Note that since we doubled the degrees of freedom, we need to be careful when calculating expectation values. Usually we only use operators with $E>0$, or, alternatively, divide by $2$ all results that we get when using the $2N$ operators.
Now we can discuss the specific case of Kitaev's chain the ground state degeneracy (your question 3). Our Hilbert space is completely spanned by the $N$ operators with $E\geq 0$. The only operator that doesn't change the energy and leaves the state in the ground state are the operators $d$ and $d^{\dagger}$ that have $E=0$ and are comprised of the edge states. Note that this gives $2$-fold degeneracy as we can either have the state $d^{\dagger}d$ empty or full, without affecting the ground state energy. Equivalently - we can arbitrarily decide if we include or exclude $E=0$ from the product that defined $|{\rm gs}\rangle$ above. This is not accidental and lies at the heart of the topology of the model.
As to your last question, I think that the point is that $\langle \bf{x} |$ should be defined properly. The ground state is not empty but is described by a many-body wave function (as in any superconductor, describing all the Cooper pairs). So thinking about this in terms of single-body Wannier functions requires re-defining everything with respect to the new ground state, and the Wannier functions correspond to the new BdG operators which are a superposition of particle and hole.
