At low energies, close to the Dirac points we can obtain the following effective Hamiltonians:
\begin{align*}
\begin{array}{ll}
\mathcal{H}_{\mathbf{K}} = \hbar v_{F}\left(\Pi_{x}\sigma^{x} + \Pi_{y}\sigma^{y}\right) & \text{at the $K$ point}
\\
\mathcal{H}_{\mathbf{K^{\prime}}} = -\hbar v_{F}\left(\Pi_{x}\sigma^{x} - \Pi_{y}\sigma^{y}\right) & \text{at the $K^{\prime}$ point}
\end{array}
\end{align*}
(where $\Pi_{i}$ is the gauge invariant momentum $\Pi_{i} = p_{i} + eA_{i}(\mathbf{r})$)
We can combine these to write:
\begin{align*}
H_{valley} &= \hbar v_{F} \left(\begin{array}{cc}\boldsymbol{\Pi}.\boldsymbol{\sigma} & 0 \\ 0 & -\boldsymbol{\Pi}.\boldsymbol{\sigma}\end{array}\right)
\\
&= \hbar v_{F}\sigma^{3}\otimes\mathbf{q}.\boldsymbol{\sigma}
\end{align*}
Here thetensor product can be thought of as valley$\otimes$sublattice, so the valley pseudospin takes the place of the usual spin we see in the massless Dirac Hamiltonian.
The spin itself is added to the Hamiltonian then, say writing it as $H_{S}$, and they (spin and valley pseudospin) are completely independent from each other. We can write the total Hamiltonian as:
\begin{align*}
H_{tot} = H_{S} \otimes H_{valley}
\end{align*}
Each spin has an associated $SU(2)$ symmetry, and due to their independent natures these combine to give an overall $SU(4)$ symmetry (since we can have entangled operators). The $SU(4)$ symmetry is generated by $\sigma^{i} \otimes \mathbb{1}$, $\mathbb{1} \otimes \sigma^{i}$,$\sigma^{i} \otimes \sigma^{j}$ for $i,j \in \{1,2,3\}$ where the first component of the tensor product is associate with the spin, and the second is the valley pseudospin.
Hope this clears some things up.
