Consider orieneted euclidean space $\mathbb{R}^3$. A vector $v\in\mathbb{R}^3$ determines a $1$-form $\omega_v^1$ where $\omega_v^1(w)=\langle v,w\rangle$, where the brackets denotes the equipped scalar product; this is an isomorphism from $\mathbb{R}^3$ to the space of $1$-forms. In the book I'm reading, it claims that we also have an isomorphism with the space of $2$-form given by $$\omega_v^2(w_1,w_2)=(v,w_1,w_2),$$ where the right hand side is the triple scalar product. Notably they remark that the isomorphisms above do not depend on the choice of oriented basis, but they both do depend on the choice of scalar product. Moreover, the second isomorphism does depends on the orientation.
I think I understand everything for the first isomorphism. For the second one, however, I think I don't really understand what the scalar triple product is. The only way I have of showing the isomorphism is by expanding via an orthonormal basis, and taking the determinant. But I think I really am missing something here because such a definition seems to completely ignore the structure of the scalar product, which the author seems to say is relevant. Is there a more "coordinate-free" description here? I am also not exactly sure what it is meant by how the second isomorphism depends on orientation.
EDIT: Does this have to do at all with a Riemannian metric? Later in the chapter the author mentions that with such a metric, we can talk about scalar, cross, and triple products. I can see where scalar products come from, but I really don't know how to understand cross products and triple products.
The triple product can simply be given as the volume given by the three vectors $v,w_1,w_2$. But given a scalar product, we can calculate this volume by taking the square root of the determinant of the Gram matrix (this is the "natural" volume form induced by the scalar product), and adjusting the sign for orientation purposes.