Let $x,y,z\in\mathbb{R}^{5}$. Let $$f(x,y,z)=2x_{2}y_{2}z_{1}+x_{1}y_{5}z_{4},\\ g(x,y)=x_{1}y_{3}+x_{3}y_{1}, \\h(w)=w_{1}-2w_{3}$$ Using Munkre's notation, if $F$ is a k-tensor, we define $AF=\sum_{\sigma}(sgn\,\,\sigma)F^{\sigma}$. Here, $\sigma$ is a permutation in the set $\{1,\dots,k\}$ and $F^{\sigma}(v_{1},\dots,v_{k})=F(v_{\sigma(1)},\dots v_{\sigma(k)})$.
Thus we can define the wedge product of the $k$-tensor $f$ with the $l$-tensor $g$ by $$f\wedge g=\frac{A(f\otimes g)}{k!l!} $$
How can I express $Af$ and $Ag$ in terms of elementary alternating tensors? And how can I express $(AF)\wedge h$ in terms of alternating elementary tensors?
If $(e_1, e_2, e_3,e_4,e_5)$ is the standard basis for $\Bbb R^5$, and $(e^1,e^2, e^3,e^4, e^5)$ is the dual basis, we have $$\begin{align} f &= 2\;e^2\otimes e^2 \otimes e^1+e^1\otimes e^5\otimes e^4 \\ g &= e^1\otimes e^3 + e^3\otimes e^1 \\ h&= e^1 - 2e^3.\end{align}$$ Now, note that by definition of $\wedge$, we have $$e^i \wedge e^j = \frac{A(e^i\otimes e^j)}{1!1!} = A(e^i \otimes e^j)\quad \mbox{and} \quad e^i\wedge e^j \wedge e^k = \frac{A(e^i\otimes e^j\otimes e^k)}{1!1!1!}= A(e^i\otimes e^j\otimes e^k).$$Since $A$ is linear, we get $$\begin{align} Af &= \require{cancel} 2\;\cancelto{0}{\color{red}{e^2}\wedge \color{red}{e^2} \wedge e^1}+e^1\wedge e^5\wedge e^4 = - e^1\wedge e^4\wedge e^5 \\ Ag &= e^1\wedge e^3 + e^3\wedge e^1 = 0 \\ Ah &= e^1 - 2e^3 = h.\end{align}$$For the last question, we just compute $$\begin{align} (Af)\wedge h &= (-e^1 \wedge e^4 \wedge e^5) \wedge (e^1 -2e^3) \\ &= -\cancelto{0}{\color{red}{e^1} \wedge e^4 \wedge e^5 \wedge \color{red}{e^1}} + 2 e^1 \wedge e^4\wedge e^5 \wedge e^3 \\ &= 2e^1 \wedge e^3 \wedge e^4 \wedge e^5.\end{align}$$