A few points:
- It is necessary to define "Riemannian volume forms" a patch at a time: you can have non-orientable Riemannian manifolds. (Symplectic manifolds are however necessarily orientable.) So you cannot just have a global construction mapping Riemannian metric to Riemannian volume form. (Consider the Möbius strip with the standard metric.)
- It is however to possible to give a definition of the Riemannian volume form locally in a way that does not depend on choosing a coordinate basis. This also showcases why there cannot be a natural map from $\mathrm{Sym}^2\to \Lambda^n$ sending inner-products to volume forms. We start from the case of the vector space. Given a vector space $V$, we know that $V$ and $V^*$ are isomorphic as vector spaces, but not canonically so. However if we also take a positive definite symmetric bilinear form $g\in \mathrm{Sym}_+^2(V^*)$, we can pick out a unique compatible isomorphism $\flat: V\to V^*$ and its inverse $\sharp: V^*\to V$. A corollary is that $g$ extends to (by abuse of notation) an element of $\mathrm{Sym}_+^2(V)$. Then by taking wedges of $g$ you get that the metric $g$ (now defined on $V^*$) extends to uniquely to a metric1 on $\Lambda^k(V^*)$. Therefore, up to sign there is a unique (using that $\Lambda^n(V^*)$ is one-dimensional) volume form $\omega\in \Lambda^n(V^*)$ satisfying $g(\omega,\omega) = 1$. But be very careful that this definition is only up to sign.
- The same construction extends directly to the Riemannian case. Given a differentiable manifold $M$. There is a natural map from sections of positive definite symmetric bilinear forms on the tangent space $\Gamma\mathrm{Sym}_+^2(T^*M) \to \Gamma\left(\Lambda^n(M)\setminus\{0\} / \pm\right)$ to the non-vanishing top forms defined up to sign. From which the usual topological arguments shows that if you fix an orientation (either directly in the case where $M$ is orientable or lifting to the orientable double cover if not) you get a map whose image now is a positively oriented volume form.
Let me just summarise by giving the punch line again:
For every inner product $g$ on a vector space $V$ there are two compatible volume forms in $\Lambda^n V$: they differ by sign. Therefore the natural mapping from inner products takes image in $\Lambda^n V / \pm$!
Therefore if you want to construct a map based on fibre-wise operations on $TM$ sending Riemannian metrics to volume forms, you run the very real risk that, due to the above ambiguity, what you construct is not even continuous anywhere. The "coordinate patch" definition has the advantage that it sweeps this problem under the rug by implicitly choosing one of the two admissible local (in the sense of open charts) orientation. You can do without the coordinate patch if you start, instead, with an orientable Riemannian manifold $(M,g,\omega)$ and use $\omega$ to continuously choose one of the two admissible pointwise forms.
1: this used to be linked to a post on MathOverflow, which has since been deleted. So for completeness: the space of $k$-tensors is the span of tensors of the form $v_1 \otimes \cdots \otimes v_k$, and you can extend $g$ to the space of $k$-tensors by setting
$$ g(v_1\otimes\cdots v_k, w_1\otimes\cdots\otimes w_k) := g(v_1, w_1) g(v_2, w_2) \cdots g(v_k, w_k) $$
and extending using bilinearity. The space $\Lambda^k(V^*)$ embeds into $\otimes^k V^*$ in the usual way and hence inherits a inner product.
I've seen in the literature the notation $C$ with some additional specifications for the contraction maps of all sorts, but the amount of decorations on the symbol $C$ varied depending on the context. See, e.g., A.Gray, Tubes, p.56, where these maps are used in the case of somewhat special tensors, and therefore the notation is simpler.
In general, there is a whole family of uniquely defined maps
$$
C^{(r,s)}_{p,q} \colon \otimes^{r}_{s} V \to \otimes^{r-1}_{s-1} V
$$
which are collectively called tensor contractions ($1 \le p \le r, 1 \le q \le s$).
These maps are uniquely characterized by making the following diagrams commutative:
$$
\require{AMScd}
\begin{CD}
\times^{r}_{s} V @> {P^{(r,s)}_{p,q}} >> \times^{r-1}_{s-1} V\\
@V{\otimes^{r}_{s}}VV @VV{\otimes^{r-1}_{s-1}}V \\
\otimes^{r}_{s} V @>{C^{(r,s)}_{p,q}}>> \otimes^{r-1}_{s-1} V
\end{CD}
$$
Explanations are in order.
Recall that the tensor products $\otimes^{r}_{s} V$ are equipped with the universal maps
$$
\otimes^{r}_{s} \colon \times^{r}_{s} V \to \otimes^{r}_{s} V
$$
where $\times^{r}_{s} V := ( \times^r V) \times (\times^s V^*)$.
Besides that, there is a canonical pairing $P$ between a vector space $V$ and its dual:
$$
P \colon V \times V^* \to \mathbb{R} \colon (v, \omega) \mapsto \omega(v)
$$
Notice that map $P$ is bilinear and can be extended to a family of multilinear maps
$$
P^{(r,s)}_{p,q} \colon \times^{r}_{s} V \to \times^{r-1}_{s-1} V
$$
by the formula:
$$
P^{(r,s)}_{p,q} (v_1, \dots, v_p, \dots, v_r, \omega_1, \dots, \omega_q, \dots, \omega_s) = \omega_q (v_p) (v_1, \dots, \widehat{v_p}, \dots, v_r, \omega_1, \dots, \widehat{\omega_q}, \dots, \omega_s)
$$
where a hat means omission.
Since maps $P^{(r,s)}_{p,q}$ are multilinear, the universal property of the maps $\otimes^{r}_{s}$ implies that there are uniquely defined maps
$$
\tilde{P}^{(r,s)}_{p,q} \colon \otimes^{r}_{s} V \to \times^{r-1}_{s-1} V
$$
and then the maps $C^{(r,s)}_{p,q}$ are given by
$$
C^{(r,s)}_{p,q} := \otimes^{r-1}_{s-1} \circ \tilde{P}^{(r,s)}_{p,q}
$$
Best Answer
To understand the trace, it is good to spell out the isomorphism between $End(V)$ and $V\otimes V^*$: $$V\otimes V^*\to End(V): v\otimes\omega \mapsto (x\mapsto \omega(x)v).$$
Under this isomorphism, composition of endomorphisms becomes $$(V\otimes V^*) \times (V\otimes V^*) \to V\otimes V^*: (v_2\otimes\omega_2, v_1\otimes \omega_1)\mapsto \omega_2(v_1)\cdot v_2\otimes \omega_1.$$
Taking the trace of this composition, one gets $\omega_2(v_1)\omega_1(v_2)$. It is then easy to see that the trace of $v_2\otimes\omega_2\circ v_1\otimes \omega_1$ is the same as that of $v_1\otimes\omega_1\circ v_2\otimes \omega_2$.