Let's first break down the symbols we know in that expression:
\begin{align}
\mathbf{\vec{S}} &= \frac{\hbar}{2}\boldsymbol{\vec{\sigma}}\tag{1}\\
&=\frac{\hbar}{2}\left(\sigma_x \boldsymbol{\hat{x}} + \sigma_y \boldsymbol{\hat{y}} + \sigma_z\boldsymbol{\hat{z}}\right)\tag{2}\\
\end{align}
$${\small
\mathbf{\vec{S}}_1\times \mathbf{\vec{S}}_2 = \left(\sigma_{y,1}\sigma_{z,2} - \sigma_{z,1}\sigma_{y,2}\right)\boldsymbol{\hat{x}} - \left(\sigma_{x,1}\sigma_{z,2} - \sigma_{z,2}\sigma_{x,1}\right)\boldsymbol{\hat{y}} + \left(\sigma_{x,1}\sigma_{y,2} - \sigma_{y,2}\sigma_{x,1}\right)\boldsymbol{\hat{z}} \!\!\!\!\tag{3}}.
$$
We now want to second-quantize all Pauli operators, and as Anyon said in this comment there's plenty of ways to do that. I mentioned two of those ways (the Jordan-Wigner transform and the Bravyi-Kitaev transform) in this answer.
The final part of the Hamiltonian which I haven't yet mentioned is $\mathbf{\vec{D}}_{12}$. In the 79-page paper you gave us, and in my brief look of it, I was only able to find this sentence about that symbol:
"$\mathbf{\vec{D}}_{12} (= - \mathbf{\vec{D}}_{21})$ depends on
details of electron wave functions and the symmetry of the crystal
structure (Moriya, 1960a; Bogdanov and Hubert, 1994)."
To second-quantize $\mathbf{\vec{D}}_{12}$ you would have to first decide what it is (depending on the details of the electron wave functions and the symmetry of the crystal structure, which suggests that it will depend on your material).
The procedure for how to second-quantize the Rashba spin-orbit coupling Hamiltonian was given in the second paper you gave us in sections IIIA and IIIB, and a similar procedure would be followed once you have enough detail about $\mathbf{\vec{D}}_{12}$.