For $v_1=ae_1+be_2$ and $v_2=ce_1+de_2$.Then in the form of wedge product
$$v_1\wedge v_2=(ad-bc)e_1\wedge e_2$$
But we also have its formula in the dual form.If we set $f\in A_k (V)$ ,and $g\in A_l (V)$.Then we have the formula: $$f\wedge g(v_1,v_2,...,v_{k+l})=\sum_{\sigma\in(k,l)-shuffles}(sgn\sigma)f(v_{\sigma (1)},...,v_{\sigma (k)})g(v_{\sigma (k)},...,v_{\sigma (k+l)}) \tag1$$
More specially for two covectors: $$f\wedge g(v_1,v_2)=f(v_1)g(v_2)-f(v_2)g(v_1)$$
Then we set $f(v_1)=a,f(v_2)=c,g(v_1)=b,g(v_2)=d$,then the result is $ad-bc$,which corresponds with the coefficients of the wedge product for vectors.
Here is my confusion:
$(Q_1)$. Is it just a coincidence? Or can we say the coincidence motivates us to define a formula like $(1)$ in the dual form?
$(Q_2)$. Does it happen in a higher dimension vector space with more vectors?That is to say,can we use formula $(1)$ to calcutae the coefficients with more vectors in a higher dimension vector space?See wiki'spage (Content 1.2):https://en.wikipedia.org/wiki/Exterior_algebra
$(Q_3)$. Can we generalize $(1)$ in a way of fixed tensor?
For $(Q_3)$,I guess it's not because we can't have a permutation or shuffles in a fixed way.
Hope I've made my exposition clear.I'm grateful to your answer,thanks.
It seems what you are missing is the fact that (for finite-dimensional spaces) there is a natural isomorphism $$\Lambda^k(V^*)\simeq \left( \Lambda^k V\right)^*$$ where your $A_k(V)$ naturally identifies with $\left( \Lambda^k V\right)^*$.
This isomorphism is given by: $$\phi_1\wedge \dots\wedge \phi_k\mapsto \left(v_1\wedge\dots\wedge v_k\mapsto \det(\phi_i(v_j))\right).$$
So the wedge product for alternating form is actually a wedge product of vectors, but in the dual space, and the "coincidence" is not actually a coincidence.
Specifically, given your $f,g\in V^*$, we have the usual wedge product in the exterior algebra $f\wedge g\in \Lambda^2(V^*)$, and using the isomorphism above we can see it as an element of $A_2(V)$ sending $(v_1,v_2)$ to $$ \begin{vmatrix} f(v_1) & f(v_2) \\ g(v_1) & g(v_2) \end{vmatrix} = ad-bc.$$