Let $S$ a oriented vector space of dimension $m$ and equipped with a inner product. Given $v \in S$, let $\omega = \phi(v) \in \mathcal{A}_{m-1}(S)$ defined by $$\omega(v_{1},...,v_{m-1}) = \langle v, v_{1},...,v_{m-1} \rangle.$$ Show that the aplication $\phi: S \to \mathcal{A}_{m-1}(S)$ is an isomorphism. If $v \in S$ is the first element of a orthonormal positive basis such that the dual is $\{ e_{1},...,e_{m} \}$ then $\phi(v) = e_{2} \wedge \dots \wedge e_{m}$. Conclude that if $\dim S = m$ then every $\omega \in \mathcal{A}_{r}(S)$ is decomposable.
Notation. $\mathcal{A}_{r}(S)$ denotes the vector space of $r$-linear forms over $S$.
Idea. First, $\dim \mathcal{A}_{m-1}(S) = \binom{m}{m-1} = m$ and, since $\phi$ is linear, we need to show that $\ker \phi = \{ 0 \}$. Note that $$\phi(v) = 0 \Longleftrightarrow \omega = 0 \Longleftrightarrow \langle v, v_{1},...v_{m-1} \rangle = 0 \Longleftrightarrow v = 0.$$ Is this right?
I couldn't prove the second part. I would like help. I didn't try to make the conclusion yet, so I wanted help only in the second part.
Your idea is indeed a correct proof; maybe in the third equivalent statement in the highlighted row one could insert "for all $v_1, \ldots, v_{m-1}$".
For the second statement, it suffices to show that the two $m-1$-forms coincide on the given orthonormal positive basis, which we denote by $v = b_1, b_2, \ldots, b_n$; but indeed,
$$ \phi(v)(b_1, \ldots, \hat{b_k}, \ldots, b_n) = \langle b_1, b_1, \ldots, \hat{b_k}, \ldots, b_n \rangle = \begin{cases} +1 & k=1 \\ 0 & \text{else}, \end{cases} $$
the same as $e_2 \wedge \cdots \wedge e_n$. Finally, let $\omega \in A_{m-1}(S)$ be arbitrary. By surjectivity of $\phi$, we find $v \in S$ so that $\phi(v) = \omega$. Then $\phi(cv) = c\omega$, where $c$ is a constant such that $\|cv\| = 1$. Then extend $v$ to an orthonormal basis by basis extension and Gram‒Schmidt and apply the second point to get that $c\omega$ is decomposable. Hence $\omega$ is decomposable by sliding in the $c$ into one of the wedge factors by multilinearity.