I have this problem in exterior algebra where I have a function B and B is defined in the following ways.
$$
B\left( \left( \begin{array}{c} u \\ v \\ w \end{array} \right), \left( \begin{array}{c} x\\y\\z \end{array} \right) \right) = uy-2uz-vx-3vz+2wx+3wz
$$
$$ B=b_{12}e^{12}+b_{13}e^{13}+b_{23}e^{23}=(b_1\otimes b_2)(e^1\otimes e^2)+(b_1\otimes b_3)(e^1\otimes e^3)+(b_2\otimes b_3)(e^2\otimes e^3) $$ where $e^1 = (1,0,0)$, $e^2 = (0,1,0)$, and $e^3 = (0,0,1)$ and I am supposed to solve for $b_1$, $b_2$, and $b_3$. If the subscript versus superscript convention is confusing, let me add that my professor denotes vectors with subscripts and linear functionals with superscripts and I too am abiding by that convention. My big issue with this is how to foil the tensor products since I do not have much experience in this area?
I'm going to write $\varepsilon^i$ instead of $e^i$ to make the notation clearer: we have $\varepsilon^i(e_j) = \delta_{ij}$. We want to write $B$ as some linear combination of the $\varepsilon^i \otimes \varepsilon^j$. Note that $(\varepsilon^i \otimes \varepsilon^j)(e_p \otimes e_q) = \delta_{ip} \delta_{jq}$ just as for the 1-dimensional case. This means if we write $$B = \sum b_{ij} \varepsilon^i \otimes \varepsilon^j$$ then $$B(e_i, e_j) = b_{ij}$$.
So we can write $B$ out explicitly with respect to the basis $e_1, e_2, e_3$: $$B = \begin{pmatrix}0 & 1 & -2 \\ -1 & 0 & -3 \\ 2 & 0 & 3 \end{pmatrix}$$ which is exactly the matrix $b_{ij}$ that you're looking for.
However, this does mean you won't get an expression of the form $B = a \varepsilon^1 \otimes \varepsilon^2 + b \varepsilon^2 \otimes \varepsilon^3 + c\varepsilon^1 \otimes \varepsilon^3$, so it's possible that I have misunderstood exactly what it is you're asking for!