I try to answer starting from the case of square matrices. There is some care to take while considering a "hidden" isomorphism of vector spaces. In any case, let $V$ be a finite dim. vector spaces over a field $\mathbb K$ (for simplicity $\mathbb R$ ), with basis $\{e_i\}$ of cardinality $n$.
It is well known that there exists an isomorphism of vector spaces
$$\Phi:\operatorname{Hom}_\mathbb K(V,V)\rightarrow V^{*}\otimes V, $$
with $$\Phi(\phi)=a_{ij}f_i\otimes e_j,$$
where $\phi\in \operatorname{Hom}_\mathbb K(V,V)$ and $\phi(e_i):=a_{ij}e_j$ for all $i,j=1,\dots,n$.
$\{f_i\}$ is the dual basis on $V^{*}$ of the basis $\{e_i\}$ on $V$, i.e. $f_i(e_j)=\delta_{ij}$.
We use the Einstein convention for repeated indices.
We know how to define the trace operator $\operatorname{Tr}$ on the space $\operatorname{Hom}_\mathbb K(V,V)$; the trace is computed on the square matrix representing each linear map in $\operatorname{Hom}_\mathbb K(V,V)$. Let us move to the r.h.s. of the isomorphism $\Phi$.
- trace operator on $V^{*}\otimes V$
Let
$$\operatorname{Tr}_1: V^{*}\otimes V\rightarrow \mathbb K, $$
be given by $\operatorname{Tr}_1(g\otimes v):=g(v)$.
Lemma $\operatorname{Tr}_1$ is linear and satisfies
$$\operatorname{Tr}_1\circ \Phi=\operatorname{Tr}.$$
proof: just use definitions.
- trace operator on $(V^{*}\otimes V)\otimes\dots\otimes (V^{*}\otimes V) $
Using the $n=1$ case we introduce
$$\operatorname{Tr}_n: \underbrace{(V^{*}\otimes V)\otimes\dots\otimes (V^{*}\otimes V)}_{n-\text{times}} \rightarrow \mathbb K, $$
with $\operatorname{Tr}_n(f_1\otimes v_1\otimes\dots\otimes f_n\otimes v_n):=\prod_{i=1}^n f_i(v_i)$.
Lemma $\operatorname{Tr}_n$ is linear and invariant under permutations on $(V^{*}\otimes V)^{\otimes n}$;
it satisfies
$$\operatorname{Tr}_n\left(\Phi(\phi_1)\otimes\dots\otimes\Phi(\phi_n)\right)=\prod_{i=1}^n \operatorname{Tr}(\phi_i), $$
for all $\phi_i\in \operatorname{Hom}_\mathbb K(V,V)$.
proof: we prove the second statement. We introduce the notation
$$\Phi(\phi_k):=
a^k_{i_kj_k}f_{i_k}\otimes e_{i_k}\in V^{*}\otimes V,$$
for all $k=1,\dots,n$.
We arrive at $$\operatorname{Tr}_n\left( (a^1_{i_1j_1}f_{i_1}\otimes e_{i_1})\otimes\dots\otimes
(a^n_{i_nj_n}f_{i_n}\otimes e_{i_n})\right)=a^1_{i_1j_1}\dots a^n_{i_nj_n}f_{i_1}(e_{i_1})\dots
f_{i_n}(e_{i_n})=\text{remember the definition of dual basis}=
a^1_{i_1j_1}\dots a^n_{i_nj_n}\delta_{i_1j_1}\dots\delta_{i_nj_n}=
a^1_{i_1i_1}\dots a^n_{i_ni_n}\\=\prod_{i=1}^n \operatorname{Tr}(\phi_i),$$
as claimed.
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
Your question indicates you should read a book on tensors or multilinear algebra, but here is a brief reply.
Tensors of type (m,n) on a vector space can be viewed as objects belonging to the space $V \otimes V \ldots \text{(m times)} V^* \otimes V^* \otimes \ldots \text{(n times)}$, where $V^*$ is the dual space to the vector space $V$ and $\otimes$ is the tensor product.
Because, by definition, dual spaces consist of one-forms that take in a vector and return a quantity in the scalar field of the vector space, there is a natural operation, called contraction, that does precisely this. So if you take an (m,n) tensor and contract one pair of indices you are left with a (m-1,n-1) type tensor. Note that the contraction inherently requires the pairing of an index representing a co-vector and an index representing a vector.
In the special case of a linear map, which is a tensor of type (1,1), the contraction operation is called a trace. It returns a scalar.
In index notation, you'd write the trace of a tensor $T^i\;_j$ as $T^i_i$.
The significance of this quantity is that it is independent of the basis you choose to find the components $T^i_j$ of the linear map T. It is an invariant function of the linear map itself. No such importance is attached to the sum of diagonal components of say a (0,2) tensor (which you might know as a quadratic form) - it would change with a change of basis.