Let $\mathbf{v}$ and $\mathbf{u}$ be vectors:
$$ \mathbf{v}:=x_v \sigma_x+ y_v \sigma_x+ z_v \sigma_v\\ \mathbf{u}:=x_u \sigma_x+ y_u \sigma_x+ z_u \sigma_v $$
What standard notation, if any, gives the following product:
$$ \mathbf{v}?\mathbf{u}=x_v(y_u+z_u)+y_v(x_u+z_u)+z_v(x_u+y_u) $$
This is an ordinary dot product; more precisely, a symmetric bilinear form with signature $(1,2)$. Consider a standard basis $\{e_1,e_2,e_3\}$ for such a form:
$$e_1\cdot e_1=1,\quad e_2\cdot e_2=-1,\quad e_3\cdot e_3=-1,\\ e_1\cdot e_2=e_1\cdot e_3=e_2\cdot e_3=0$$
The null cone, the set of all vectors squaring to $0$, has $e_1$ as its axis. Now define a set of vectors $\{\sigma_k\}$ equally spaced around the cone:
$$\sigma_1=\frac{\sqrt2}{\sqrt3}\big(e_1+e_2\big),\quad\sigma_2=\frac{\sqrt2}{\sqrt3}\Big(e_1-\frac12e_2+\frac{\sqrt3}{2}e_3\Big),\quad\sigma_3=\frac{\sqrt2}{\sqrt3}\Big(e_1-\frac12e_2-\frac{\sqrt3}{2}e_3\Big)$$
These are linearly independent, so $\{\sigma_k\}$ can be used as a basis instead of $\{e_k\}$. You can calculate $\sigma_1\cdot\sigma_1=\sigma_2\cdot\sigma_2=\sigma_3\cdot\sigma_3=0$ , and $\sigma_1\cdot\sigma_2=\sigma_1\cdot\sigma_3=\sigma_2\cdot\sigma_3=1$ . So the dot product of any two vectors is
$$(u_1\sigma_1+u_2\sigma_2+u_3\sigma_3)\cdot(v_1\sigma_1+v_2\sigma_2+v_3\sigma_3)=u_1v_2+u_1v_3+u_2v_1+u_2v_3+u_3v_1+u_3v_2$$
which agrees with your formula, up to differences in notation.