TL;DR: The given form of the channel is essentially an obscured way of writing down the channel's Pauli transfer matrix. One set of Kraus operators coincides (up to vectorization) with the eigenvectors of the Choi matrix. Therefore, modulo rearrangements of matrix elements, the task amounts to the conversion from the Pauli transfer matrix to the Choi matrix followed by eigendecomposition.
Affine transformation of the Bloch sphere to Pauli transfer matrix
By considering the action of $\Phi(\rho) = \frac{1}{2}\left(I+(T\vec{r}+\vec{t})\cdot\vec{\sigma}\right)$ on each Pauli operator, we see that the channel's Pauli transfer matrix is
$$
R(\Phi)=\begin{bmatrix}1 & 0\\
\vec{t}&T
\end{bmatrix}.\tag1
$$
Pauli transfer matrix to Choi matrix
Let's denote the three columns of $T$ with $\vec{t}_X$, $\vec{t}_Y$, and $\vec{t}_Z$ and $\vec{t}$ with $\vec{t}_I$. We compute the channel's Choi matrix as
$$
\begin{align}
J(\Phi)&:=(\Phi\otimes\mathcal{I})(|\psi\rangle\langle\psi|)\tag2\\
&=(\Phi\otimes\mathcal{I})\left(\frac12\sum_{P\in\{I,X,Y,Z\}}P\otimes P^T\right)\tag3\\
&=\frac12\sum_{P\in\{I,X,Y,Z\}}\Phi(P)\otimes P^T\tag4\\
&=\frac{I\otimes I}{2}+\sum_{P\in\{I,X,Y,Z\}}\frac{\vec{t}_P\cdot\vec{\sigma}}{2}\otimes P^T\tag5
\end{align}
$$
where $|\psi\rangle=|00\rangle+|11\rangle$ is an unnormalized maximally entangled state of two qubits.
Choi matrix to a Kraus representation
Next, we rewrite the Choi matrix as a sum of rank-one matrices
$$
J(\Phi)=\sum_k|\psi_k\rangle\langle\psi_k|.\tag6
$$
This can be accomplished for example by computing the eigendecomposition of $J(\Phi)$. Finally, by rearranging the four components of $|\psi_k\rangle$ we construct the matrix $A_k$ such that $|\psi_k\rangle=(A_k\otimes I)|\psi\rangle$. But then
$$
J(\Phi)=\sum_k(A_k\otimes I)|\psi\rangle\langle\psi|(A_k^\dagger\otimes I)\tag7.
$$
Comparing to $(2)$ we conclude that $\Phi(\rho)=\sum_kA_k\rho A_k^\dagger$ is a Kraus representation of $\Phi$.
