Let $n\geq4 $ and $ \omega \in \Lambda^2(\mathbb R^n)^* $ an alternating 2-linear form. How can one show that for $v_1,\ldots,v_4 \in \mathbb R^n$:
$$ \omega \land \omega(v_1,\ldots, v_4)=2 \omega(v_1,v_2)\omega(v_3,v_4)-2\omega(v_1,v_3)\omega(v_2,v_4)+2\omega(v_1,v_4)\omega(v_2,v_3).$$
Now I thought to try something by using the alternator, but still I cant solve this. I got something like $$\sum sign(\sigma)\omega(v_{\sigma(1)},v_{\sigma(2)})\omega(v_{\sigma(3)},v_{\sigma(4)})+\sum sign(\sigma)\omega(v_{\sigma(1)},v_{\sigma(2)})\omega(v_{\sigma(3)},v_{\sigma(4)})+\sum sign(\sigma)\omega(v_{\sigma(1)},v_{\sigma(2)})\omega(v_{\sigma(3)},v_{\sigma(4)}) .$$ I am not sure about my idea. Any help or a hint is much appreciated. Thanks!
You might find it useful to prove the following formula for the wedge product:
$$ (\omega \wedge \eta)(v_1,v_2,v_3,v_4) = \sum_{\sigma \in S_{2,2}} \operatorname{sign}(\sigma) \omega(v_{\sigma(1)}, v_{\sigma(2)})\eta (v_{\sigma(3)},v_{\sigma(4)}) $$
where $S_{2,2}$ denotes the $(2,2)$-shuffles which are the permutations $\sigma$ of $\{1,2,3,4\}$ that satisfy $\sigma(1) < \sigma(2)$ and $\sigma(3) < \sigma(4)$. The advantage of this formula (which can be proved by a combinatorical argument and generalizes for higher order forms) is that it involves working with only $|S_{2,2}| = {4 \choose 2} = 6$ permutations and not $4! = 24$ permutations.
The relevant permutations in your case are
$$ \operatorname{id}, (23), (243), (123), (1243), (13)(24) $$
and so
$$ (\omega \wedge \omega)(v_1,v_2,v_3,v_4) = \omega(v_1,v_2)\omega(v_3,v_4) - \omega(v_1,v_3)\omega(v_2,v_4) + \omega(v_1,v_4)\omega(v_2,v_3) \\ + \omega(v_2,v_3)\omega(v_1,v_4) - \omega(v_2,v_4)\omega(v_1,v_3) + \omega(v_3,v_4)\omega(v_1,v_2) \\ = 2\omega(v_1,v_2)\omega(v_3,v_4) - 2\omega(v_1,v_3)\omega(v_2,v_4) + 2\omega(v_2,v_3)\omega(v_1,v_4).$$
Alternatively, you can brute-force your way through all $4! = 24$ permutations and use the fact that $\omega$ is alternating to get the same expression. More explicitly, you have
$$ (\omega \wedge \eta)(v_1,v_2,v_3,v_4) = \frac{4!}{2!\cdot 2!} \operatorname{Alt}(\omega \otimes \eta)(v_{\sigma(1)}, v_{\sigma(2)}, v_{\sigma(3)}, v_{\sigma(4)}) \\ = \frac{1}{2! \cdot 2!} \sum_{\sigma \in S_{4}} \operatorname{sign}(\sigma) \omega(v_{\sigma(1)}, v_{\sigma(2)})\eta (v_{\sigma(3)},v_{\sigma(4)}). $$
I have already computed the right hand side for six permutations. Now show that each of the six permutations above comes with three permutations such that result in the same value. For example, the permutations
$$ \operatorname{id}, (12)(34), (12), (34) $$
fix $\{ 1,2 \}$ and $\{ 3, 4 \}$ and give us
$$ \omega(v_1,v_2)\omega(v_3,v_4) + \omega(v_2,v_1)\omega(v_4,v_3) - \omega(v_2,v_1)\omega(v_3,v_4) - \omega(v_1,v_2)\omega(v_4,v_3) = 4 \omega(v_1,v_2) \omega(v_3, v_4) $$
and similarly for each of the five other permutations.