Geometric product of 2 vectors derived from Pauli matrices 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.
 A: Using 3 dimensional vectors rather then the 2 dimensional ones in the original question we get:
$
\mathbf{\vec{u}} = u_1 {\color{blue}\sigma_1} + u_2 {\color{blue}\sigma_2} + u_3 {\color{blue}{\sigma_3}}$
$\mathbf{\vec{v}} = v_1 {\color{blue}\sigma_1} + v_2 {\color{blue}\sigma_2} + v_3 {\color{blue}\sigma_3}$
$
\begin{align}
\mathbf{\vec{u}\vec{v}} &= 
\left( \begin{array}{ll}
  u_3       & \phantom{-}u_1-u_2 i  \\
  u_1+u_2 i & -u_3  
\end{array} \right)
\left( \begin{array}{ll}
  v_3       & \phantom{-}v_1-v_2 i  \\
  v_1+v_2 i & -v_3  
\end{array} \right)      \\ &=
\left( \begin{array}{rr}
     (u_3)(v_3) + (u_1-u_2 i)(v_1+v_2 i)    &  (u_3)(v_1-v_2 i) + (u_1-u_2 i)(-v_3) \\
     (u_1+u_2 i)(v_3) + (-u_3 )(v_1+v_2 i)  &  (u_1+u_2 i)(v_1-v_2 i) + (-u_3 )(-v_3)
\end{array} \right)    \\ &=
\left( \begin{array}{rr}
  u_1 v_1 + u_2 v_2 + v_3 u_3  + (u_1 v_2 - u_2 v_1)i     &   u_3 v_1 - u_1 v_3 + (u_2 v_3 - u_3 v_2)i  \\
  u_1 v_3 - u_3 v_1 + (u_2 v_3 - u_3 v_2)i                &   + u_1 v_1 + u_2 v_2 + v_3 u_3 + (u_2 v_1 - u_1 v_2)i
\end{array} \right)    \\ &=
(u_1 v_1 + u_2 v_2 + v_3 u_3) \cdot {\color{red} I_2} 
+ (u_1 v_2 - u_2 v_1) \cdot {\color{green} {\boldsymbol{\hat{\imath}}}} 
+ (u_3 v_1 - u_1 v_3) \cdot {\color{green} {\boldsymbol{\hat{\jmath}}}} 
+ (u_2 v_3 - u_3 v_2) \cdot {\color{green} {\boldsymbol{\hat{k}}}}      \\ &
\text{Now switching back to ordinary vector notation} \\ &=
\mathbf{\vec{u}} \cdot \mathbf{\vec{v}} + \mathbf{\vec{u}} \wedge \mathbf{\vec{v}} =
\left( \begin{array}{rr}
    (\mathbf{\vec{u}} \cdot \mathbf{\vec{v}})    &  {\color{green}{(u_1 v_2 - u_2 v_1)}}           &  (u_1 v_3 - u_3 v_1)  \\
    (u_2 v_1 - u_1 v_2)   &  (\mathbf{\vec{u}} \cdot \mathbf{\vec{v}})            &  {\color{green}{(u_2 v_3 - u_3 v_2)}}  \\
    {\color{green}{(u_3 v_1 - u_1 v_3)}}   &  (u_3 v_2 - u_2 v_3) &  (\mathbf{\vec{u}} \cdot \mathbf{\vec{v}})
\end{array} \right)   
\end{align}$
