( I copied some text from a previous answer of mine)
Defining the Choi and $\chi$ matrix
The Choi matrix is a direct result of the Choi-Jamiolkowski isomorphism. Some intuition on what this is can be found in this previous answer by Norbert Schuch. Consider the maximally entangled state $|\Omega \rangle = \sum_{\mathrm{i}}|\mathrm{i}\rangle \otimes |\mathrm{i}\rangle$, where $\{|\mathrm{i}\rangle\}$ forms a basis for the space on which $\rho$ acts. (Note that we thus have a maximally entangled state of twice as many qubits).
The Choi matrix is the state that we get when on one of these subsystems $\Lambda$ is applied (leaving the other subsystem intact):
\begin{equation}
\rho_{\mathrm{Choi}} = \big(\Lambda \otimes I\big) |\Omega\rangle\langle\Omega|.
\end{equation}
As the Choi matrix is a state, it must be positive semidefinite (corresonding the the CP constraint) and must have unit trace (necessary but not sufficient for the TP constraint).
The process- or $\chi$-matrix comes from the fact that we can write our map as a double sum:
\begin{equation}
\Lambda(\rho) = \sum_{m,n} \chi_{mn}P_{m}\rho P_{n}^{\dagger},
\end{equation}
where $\{P_{m}\}$ & $\{P_{n}\}$ form a basis for the space of density matrices; we use the Pauli basis $\{I,X,Y,Z\}^{\otimes n}$ (thereby omitting the need for the $\dagger$ at $P_{n}$). The matrix $\chi$ now encapsulates all information of $\Lambda$; the CP constraint reads that $\chi$ must be positive semidefinite, and the trace constraint reads that $\sum_{m,n}\chi_{mn}P_{n}P_{m} \leq I$ (with equality for TP).
Computing one from another
From this, we get the following two identities:
\begin{equation}
\begin{split}
\rho_{\mathrm{Choi}} &= \sum_{m,n} \chi_{m,n} |P_{m}\rangle\rangle\langle\langle P_{n}|, \\
\chi_{m,n} &= \langle\langle P_{m} | \rho_{\mathrm{Choi}} |P_{n}\rangle\rangle,
\end{split}
\end{equation}
where $|P_{m}\rangle\rangle$ is the 'vectorized' version of $P_{m}$, which is essentially just the columns of $P_{m}$ stacked on top of each other, giving a vector. That answers question 3.
Again I shamelessly 'self-promote': in the first appendix of my thesis I work through proofs of all these relations. The most intuitive way is by using the Kraus decomposition as an intermediary, but it is not needed.
Relationship between the two
From this, you can see that the Choi matrix and the chi matrix do indeed have some relationship. In fact, by choosing either the (qubit)-basis in which we express the Choi matrix, or choosing the (operator)-basis that we associate with the $\chi$-matrix, they can be one and the same.
As @AdamZalcman has pointed out in his comment (Thank you!), from the identity $\chi_{m,n} = \langle \langle P_{m}|\rho_{\mathrm{Choi}}| P_{n}\rangle\rangle$ we can choose the $P_{m/n}$ so that we just select the $m$-th row and $n$-th column of $\rho_{\mathrm{Choi}}$. This works if $P_{k} = |i\rangle \langle j|$, with $k = id + j$. Since both $i$ and $j$ run from $0$ to $d-1$ (indicating the column and row, respectively), this gives exactly $d^{2}$ elements.
The same effect can be reached if one expresses the Choi matrix in a different basis, while keeping the $P_{k}$ associated with $\chi_{m,n}$ the usual Paulis. For the two to coincide then (i.e. $\chi_{m,n} = \rho_{\mathrm{Choi}}^{m,n}$), we see that $\rho_{\mathrm{Choi}}^{m,n}$ should be expressed in the `vectorized-Pauli-basis' (which is a set of states, i.e. a basis for the Hilbert space!) - this is exactly the Bell basis.
