Let $V,W$ be real vector spaces of dimension $d$, and let $T \in \text{Hom}(V,W)$.
Is there a "natural piece of additional information" required in order to give a meaningful interpretation of the trace of $T$? (which is "less than" choosing an isomorphism $V \cong W$).
Let me explain a bit what do I mean by "additional information" through other examples:
Suppose we want to define the determinant of $T$. Then sufficient additional information is a choice of preferred volume forms on $V,W$. Assuming we are given such forms, we can then define $\det T$ by requiring $$T^*(\text{Vol}_W)=\det T \cdot \text{Vol}_V.$$
Perhaps an even more economical option would be to provide an isomorphism $\bigwedge^d V \simeq \bigwedge^d W$. Since $\bigwedge^d T:\bigwedge^d V \to \bigwedge^d W $, by composing it with the given isomorphism we can identify $\bigwedge^d T$ with a map $\bigwedge^d V \to \bigwedge^d V$ which can then be naturally identified with a scalar.
(Choosing a volume form is equivalent to choosing an isomorphism $\bigwedge^d V \cong \mathbb{R}$).
Of course, we could be given an isomorphism $V \simeq W$ and consider $T$ as a map $V \to V$, but my point is that this is much more than we really need.
In the same spirit, in order to define the singular values of a map, we need to choose inner products on both spaces. (Which is again less information then choosing isomorphisms of $V,W$ to $\mathbb{R}^d$).
The question is whether there is an "analogous" piece of extra structure which naturally fits into the definition of the trace. Of course, when $V=W$ (or when we are given some isomorphism between $V$ and $W$) we can define the trace of $T$ as a map $V \to V$.
(A coordinate-free version would be using the natural identification $V^* \otimes V \cong \text{Hom}(V,V)$, and the contraction map $V^* \otimes V \to F$. Alternatively we can choose a basis, and use the fact the trace of a matrix is invariant under conjugation).
I am aware that it's not entirely clear whether the trace can be given a purely "geometric meaning" (in the sense that it's not invariant under isometries for instance). This is different than the situation with the determinant, or the singular values.
Edit:
In a slight contrast to the last contrast, here is some evidence that the trace is at least "partially" a geometric creature: The space of trace-free matrices is the tangent space at the identity to $SL_d$ (the group of volume preserving matrices). Thus, I am guessing that we shall need at least something like an isomorphism $\bigwedge^d V \simeq \bigwedge^d W$ (in order to define a notion of volume preservation for maps). Using this isomorphism, we can now identify $\bigwedge^d T$ as a map $\bigwedge^d V \to \bigwedge^d V$, and define $$ "SL"=\{ T \in \text{Hom}(V,W) \, | \, \bigwedge^d T=\text{Id}_{\bigwedge^d V} \}. $$
The problem is that I am not sure if we can "single out" which element is the analog of the "identity map" among all these elements of our $"SL"$. Thus, I am not sure that a choice of an isomorphism $\bigwedge^d V \simeq \bigwedge^d W$ would suffice to even define what is "trace=$0$"...
As you observe, if we choose an isomorphism $A:W\to V$, we can define the trace of $f:V\to W$ to be $$\mathrm{Tr}_A(f):=\mathrm{Tr}(fA)=\mathrm{Tr}(A f).$$ If $B:W\to V$ is another isomorphism, then $B=MA$ for some $M\in GL(V)$, and $$ \mathrm{Tr}_B(f)=\mathrm{Tr}(MA f).$$ If $\mathrm{Tr}_A=\mathrm{Tr}_B$, then $\mathrm{Tr}(MC)=\mathrm{Tr}(C)$ for every $C\in\mathrm{End}(V)$. This implies $\mathrm{Tr}((M-\mathrm{Id}_V)C)=0$ for all $C$, which implies $M=\mathrm{Id}_V$. So the trace function determines the isomorphism from $W$ to $V$.
By contrast, $det(MA)=det(A)$ for all $A$ if and only if $det(M)=1$. So knowing the determinant function only defines an isomorphism of $W$ with $V$ modulo determinant $1$ maps. The orbit space $\mathrm{Iso}(W,V)/SL(V)$ can be identified with $\mathrm{Iso}(\bigwedge^d W,\bigwedge^d V)$.