The 3 Pauli matrices are:
${\color{blue}{\sigma_1}}$ = $ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
${\color{blue}{\sigma_2}}$ = $ \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$
${\color{blue}{\sigma_3}}$ = $ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
Define 2 "vectors" to be:
$ \mathbf{\vec{u}} = u_1 {\color{blue}\sigma_1} + u_3 {\color{blue}\sigma_3}= \begin{pmatrix} u_3 & u_1 \\ u_1 & -u_3 \end{pmatrix}$
$ \mathbf{\vec{v}} = v_1 {\color{blue}\sigma_1} + v_3 {\color{blue}\sigma_3} = \begin{pmatrix} v_3 & v_1 \\ v_1 & -v_3 \end{pmatrix}$
It then follows that:
$ \begin{align} \mathbf{\vec{u}\vec{v}} &= \left( \begin{array}{rr} u_3 & u_1 \\ u_1 & -u_3 \end{array} \right) \left( \begin{array}{rr} v_3 & v_1 \\ v_1 & -v_3 \end{array} \right) \\ &= \left( \begin{array}{rr} u_1 v_1 + u_3 v_3 & u_3 v_1 - u_1 v_3 \\ u_1 v_3 - u_3 v_1 & u_1 v_1 + u_3 v_3 \end{array} \right) \\ &= (u_1 v_1 + u_3 v_3) \cdot {\color{red} I_2} + (u_3 v_1 - u_1 v_3 ) \cdot {\color{green} {\boldsymbol{\hat{\jmath}}}} \\ &= \mathbf{\vec{u}} \cdot \mathbf{\vec{v}} + \mathbf{\vec{u}} \wedge \mathbf{\vec{v}} \end{align}$
Where:
${\color{green} {\boldsymbol{\hat{\jmath}}}} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = quaternion $
${\color{green} {\boldsymbol{\hat{\jmath}}}}$ is rotation from $\sigma_3$ to $\sigma_1$
Edit: Sorry. I got my u's and v's mixed up and got a little confused with the wedge product. The solution to the problem is that there was never any problem to begin with.
