Let's assume a fully polarized [Stokes vector][1], i.e. $I=p=1$. The last three components of the Stokes vector, $(S_1,S_2,S_3)$, are then parametrized according to
\begin{equation}
\vec S=\pmatrix{
\cos 2\psi\cos2\chi\\
\sin 2\psi\cos2\chi\\
\sin 2\chi\tag{1}
}
\end{equation}
In the $HV$ basis where $H=(1,0)^T, D=\tfrac 1{\sqrt{2}}(1,1)^T$ and $R=\tfrac 1 {\sqrt 2}(1, -i)^T$, a Jones Vector that reproduces this Stokes vector is given by
\begin{align}
J=\pmatrix{
\cos\psi\cos\chi+i\sin\psi\sin\chi\\
\sin\psi\cos\chi-i\cos\psi\sin\chi\tag{2}
}
\end{align}
This Jones vector is of course up to a global phase factor. It seems that this expression could be simplified further but I could not find a nicer expression.

To go the other way around, start with an arbitrary Jones vector
\begin{align}
J=\pmatrix{
\cos\theta e^{i\alpha}\\
\sin\theta\tag{3}
}
\end{align}
Then the corresponding Stokes vector is given by
\begin{align}
\vec S=\pmatrix{
\cos 2\theta\\
\sin2\theta \cos\alpha\\
\sin2\theta \cos\alpha
}\tag{4}
\end{align}

Finally, we can express these quantities using the density operator, which allows us to connect the degree of polarization $p$ more closely to the Jones vector. The density matrix for a pure state is given by
\begin{align}
\rho=|\psi\rangle\langle \psi |\tag{5}
\end{align}
while for a mixed state it is given by
\begin{align}
\rho=\sum_i p_i|\psi_i\rangle\langle \psi_i |\tag{6}
\end{align}
where $\sum_i p_i=1$ and $\psi_i$ form an orthonormal basis.

For a general Jones vector, this matrix is given by
\begin{align}
\rho=\left(
\begin{array}{cc}
 \cos ^2(\theta ) & e^{i \alpha } \sin (\theta ) \cos (\theta ) \\
 e^{-i \alpha } \sin (\theta ) \cos (\theta ) & \sin ^2(\theta ) \\
\end{array}
\right)\tag{7}
\end{align}
To find the density matrix for the Stokes vector, we have to create a mixed and we have to define an orthonormal basis first. We can easiliy construct a vector that is orthogonal to (2) using
\begin{align}
J_\perp=\pmatrix{-J_2^*\\J_1^*}=\pmatrix{
-\sin\psi\cos\chi-i\cos\psi\sin\chi\\
\cos\psi\cos\chi-i\sin\psi\sin\chi
} \tag{8}
\end{align}
Finally, the density matrix of the Stokes vector is given by
\begin{align}
\rho=(\tfrac 1 2+\tfrac{p}{2})|J\rangle\langle J|+(\tfrac 1 2-\tfrac{p}{2})|J_\perp\rangle\langle J\perp|
\end{align}

\begin{align}
\rho=\left(
\begin{array}{cc}
 \frac{1}{2} (1+p \cos (2 \xi ) \cos (2 \psi )) & \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )+i \sin (2 \xi )) \\
 \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )-i \sin (2 \xi )) & \frac{1}{2} (1-p \cos (2 \xi ) \cos (2 \psi )) \\
\end{array}
\right)
\tag{9}
\end{align}

To reconstruct the Stokes vector, we can use$^1$:
\begin{align}
S_i=S_0\mathrm{Tr}\left( \rho \,\hat\sigma_i \right)
\end{align} 
where
\begin{align}
&\hat\sigma_0=1\!\!1, &&\hat\sigma_1=|R\rangle\langle L|+|L\rangle\langle R|, \\
&\hat\sigma_2=i(|R\rangle\langle L|-|L\rangle\langle R|), &&\hat\sigma_3=|R\rangle\langle R|-|L\rangle\langle L|
\end{align}
Note that $\hat \sigma_2$ has a minus sign compared to the source$^1$. This is because the authors use an alternative definition of $D$.

Alternatively, we could construct a density matrix straight from the Stokes parameters using
\begin{align}
\rho=\frac 1 2\sum_{i=0}^3 \frac{S_i}{S_0}\hat\sigma_i
\end{align} 

Disclaimer: I derived a lot of these formulas myself. Although I checked them with Mathematica, if you want to use these for anything serious you might want to check them yourself.

1: [https://doi.org/10.48550/arXiv.quant-ph/0103121][2]


  [1]: https://en.wikipedia.org/wiki/Stokes_parameters#Definitions
  [2]: https://%20https://doi.org/10.48550/arXiv.quant-ph/0103121