Both bivectors and $(0,2)$-tensors are mathematical structures that take in $2$ vectors and produce a scalar. Similar as in this prior post I wrote, I would like to dumb down the mechanics of these operations with a simple example.
The goal is to understand at a very immediate level why
Let's say that we have two covectors: $\beta =\begin{bmatrix}\sqrt \pi &\sqrt[3]\pi\end{bmatrix}$ and $\gamma=\begin{bmatrix}\frac 1 3&\frac 15\end{bmatrix},$ which will form the coefficients for the wedge and tensor products.
The wedge product is $\beta \wedge \gamma$
$$\begin{align} \beta \wedge \gamma&= \sqrt \pi \frac 13 \;e^1\wedge e^1 + \sqrt\pi \frac 15 \;e^1 \wedge e^2 + \sqrt[3]\pi\frac 13\; e^2 \wedge e^1 + \sqrt[3]\pi \frac 15 e^2\wedge e^2 \\[2ex] &= 0 + \sqrt\pi \frac 15 \;e^1 \wedge e^2 + \sqrt[3]\pi\frac 13\; e^2 \wedge e^1 + 0 \\[2ex] &=\left( \sqrt \pi \frac 15 - \sqrt[3]\pi \frac 13 \right) \; e^1\wedge e^2\tag 2 \end{align}$$
If we feed two vectors to this form, say $\vec v=\begin{bmatrix} 2&3\end{bmatrix}^\top$ and $\vec w=\begin{bmatrix}4&5 \end{bmatrix}^\top$ we end up with
$$\begin{align} \left(\left( \sqrt \pi \frac 15 - \sqrt[3]\pi \frac 13 \right) \; e^1\wedge e^2\right)[\vec v, \vec w] &=\left( \sqrt \pi \frac 15 - \sqrt[3]\pi \frac 13 \right) \; \det\begin{bmatrix}2&4\\3&5 \end{bmatrix}\\[2ex] &=\left( \sqrt \pi \frac 15 - \sqrt[3]\pi \frac 13 \right)\left(2\cdot 5-3\cdot 4\right) \end{align}\tag 3$$
Now coparing to the tensor product $\beta\otimes\gamma$:
$$\beta \otimes \gamma =\begin{bmatrix} \sqrt \pi \frac 13 \; e^1\otimes e^1 & \sqrt \pi \frac 1 5 \;e^1\otimes e^2\\ \sqrt[3]\pi\frac 1 3 \, e^2\otimes e^1 & \sqrt[3]\pi \frac 1 5\,e^2\otimes e^2 \end{bmatrix}$$
Feeding the vectors $\vec v$ and $\vec w$ first as $\left(\beta \otimes \gamma\right)[\vec v, \vec w],$ followed by $\left(\beta \otimes \gamma\right)[\vec w, \vec v]:$
$$\begin{align} \left(\beta \otimes \gamma\right)[\vec v, \vec w]&= \begin{bmatrix}2&3\end{bmatrix} \begin{bmatrix} \sqrt \pi \frac 13 \; e^1\otimes e^1 & \sqrt \pi \frac 1 5 \;e^1\otimes e^2\\ \sqrt[3]\pi\frac 1 3 \, e^2\otimes e^1 & \sqrt[3]\pi \frac 1 5\,e^2\otimes e^2 \end{bmatrix} \begin{bmatrix}4\\5\end{bmatrix}\\[2ex] &=\sqrt \pi \frac 13 4\cdot 2 + \sqrt \pi \frac 15 5\cdot 2+\sqrt[3]\pi \frac 13 4\cdot 3+\sqrt[3]\pi\frac 15 5 \cdot 3 \end{align}$$
And
$$\begin{align} \left(\beta \otimes \gamma\right)[\vec w, \vec v]&= \begin{bmatrix}4&5\end{bmatrix} \begin{bmatrix} \sqrt \pi \frac 13 \; e^1\otimes e^1 & \sqrt \pi \frac 1 5 \;e^1\otimes e^2\\ \sqrt[3]\pi\frac 1 3 \, e^2\otimes e^1 & \sqrt[3]\pi \frac 1 5\,e^2\otimes e^2 \end{bmatrix} \begin{bmatrix}2\\3\end{bmatrix}\\[2ex] &=\sqrt \pi \frac 13 2\cdot 4 + \sqrt \pi \frac 15 3\cdot 4+\sqrt[3]\pi \frac 13 2\cdot 5+\sqrt[3]\pi\frac 15 3 \cdot 5 \end{align}$$
The difference is therefore:
$$\left(\beta \otimes \gamma\right)[\vec v, \vec w]-\left(\beta \otimes \gamma\right)[\vec w, \vec v]=\left(\sqrt \pi \frac 15-\sqrt[3]\pi\frac 13\right)\left(2\cdot 5 - 3 \cdot 4 \right)\tag 4$$
Now, (4) is identical to (3). The question is, then, about the $\frac 12$ factor to fulfill equation (1)? Why is there no need to divide (4) by $2$ to fulfill equation (1)?