We show here how any interpretation of $\operatorname{Tr} A$ when $A : V \to V$ is an isomorphism can be extended to an interpretation of the trace of an arbitrary endomorphism.

As described elsewhere, if you view $A : V \to V$ as a vector field on $V$ in the canonical way then the trace of $A$ is the same as its divergence so in the case where $A$ is an isomorphism there is a pleasing geometric interpretation readily available, which I'll assume that you're comfortable with. However, this interpretation is not satisfactory when $A$ is not surjective, as shown by this example:

If $A : \mathbb{R}^3 \to \mathbb{R}^3$ is such that $\operatorname{Im} A$ is $2$-dimensional and $A$ is volume-increasing (i.e. $\operatorname{div}(A) = \operatorname{Tr}(A) > 0$) then $A$ takes a bucket of $3$-d water (i.e. a subset of $\mathbb{R}^3$) and ''compresses it down'' into a $2$-d ''paper'' (i.e. into $\operatorname{Im} A$); but it is not clear (at least to me) how anyone could be expected to say that $A$ has increased the volume of this bucket of water simply because the quantity $\operatorname{div}(A) = \operatorname{Tr}(A)$ happens to be positive!

Nevertheless, the equality $\operatorname{div}(A) = \operatorname{Tr}(A)$ is our best bet at finding a geometric interpretation (Spoiler: we should look at a particular vector subspace of $V$). To begin, let
$$V^{(0)} = \operatorname{domain}(A) = V,\;\; V^{(i+1)} = A\left(V^{(i)}\right),\; \textrm{ and }d^i = \dim V^{(i)}$$
so that $V^{(1)} = \operatorname{Im} A = A\left(V^{(0)}\right)$, $V^{(i+1)} \subseteq V^{(i)}$, and $d^{i+1} \leq d^i$. Let $N \geq 0$ be the smallest integer s.t. $d^{N+1} = d^N$ and denote this common value by $d$. Let $W := V^{(N)}$. As discussed above, we may assume that $N \geq 1$ (i.e. that $A$ is not surjective) although this is not necessary.

To cut to the chase, what is shown after this paragraph is that the restriction $A\big\vert_W : W \to W$ of $A$ onto $W := V^{(N)}$ is an isomorphism. Furthermore, $\operatorname{Tr}(A) = \operatorname{Tr}\left(A\big\vert_W\right)$ and it will be clear that $W$ is the unique largest vector subspace $S$ of $V$ on which $A$ restricts to an isomorphism $A\big\vert_S : S \to S$. All of this allows us to conclude that to geometrically interpret $\operatorname{Tr}(A)$, one should restrict their focus to geometrically interpreting the isomorphism $A\big\vert_W : W \to W$ rather than $A : V \to V$ itself. This isn't surprising since just as the trace of a matrix does not depend on the "elements off the diagonal", so too does the geometric interpretation of trace not depend on the "space off of $W$."

We now prove the above claim. Inductively construct a basis $\left(e_1, \dots, e_{\dim V}\right)$ for $V$ such that for all $i \geq 0$, $\left(e_1, \dots, e_{d^i}\right)$ is a basis for $V^{(i)}$. Let $\left(\varepsilon^1,\dots, \varepsilon^{\dim V}\right)$ be the dual basis of $e_{\bullet}$ and note in particular that:
$$\textrm{(1) whenever }d^{i + 1} < l \leq d^i\textrm{ then }\varepsilon^l\textrm{ vanishes on }V^{(i + 1)}.$$

Since $(e_1, \dots, e_{d^1})$ is a basis for the range of $A$ we may, for any $v \in V^{(0)},$ write
$$A(v) = \varepsilon^1(A(v)) e_1 + \cdots + \varepsilon^{d^1}(A(v)) e_{d^1}$$
so that $A = (\varepsilon^l \circ A) \otimes e_l$ (the sum ranging over $l = 1, \dots, d^1$) and hence
$$\operatorname{Tr}(A) = (\varepsilon^l \circ A)(e_l) = \varepsilon^1(A(e_1)) + \cdots + \varepsilon^{d^1}\left(A\left( e_{d^1} \right)\right)$$
which shows that $\operatorname{Tr}(A)$ actually depends only on the range of $A$ (i.e. $V^{(1)}$).
Now since $e_1, \dots, e_{d^1}$ are (by definition) in $V^{(1)}$, all of $A\left(e_1\right), \dots, A\left(e_{d^1}\right)$ belong to $A\left(V^{(1)}\right) = V^{(2)}$ so that from $(1)$ it follows that
$$\operatorname{Tr}(A) = \varepsilon^1\left(A\left(e_1\right)\right) + \cdots + \varepsilon^{d^2}\left(A\left( e_{d^2} \right)\right)$$

Continuing this inductively $N \leq \dim V$ times shows that
$$\operatorname{Tr}(A) = \varepsilon^1\left(A\left(e_1\right)\right) + \cdots + \varepsilon^{d}\left(A\left(e_d\right)\right)$$
so that $\operatorname{Tr}(A)$ depends *only* on $W = V^{(N)}$.
Since by definition of $N$, the map $A\big\vert_W : W \to W$ is surjective, it is an isomorphism and furthermore, it should be clear that $W$ is the unique largest subspace of $V$ on which $A$ restricts to an isomorphism. QED

To summarize, going back to the divergence ''bucket of water'' example above, in the case where $A : V \to V$ is an arbitrary linear map we can imagine being given some initial ''bucket of water'' $V = V^{(0)}$ and then imagine $A$ as repeatedly (and eternally) deforming this same water until eventually (i.e. after $N$ iterations) $A$ would have ''pushed'' or ''compressed'' all of $V$ onto some vector subspace $W = V^{(N)}$ (which is also the unique the largest subspace $W$ of $V$ that $A$ maps back onto itself) i.e. all of $V$ would eventually ''flow into'' $W$.
It is at this point that $A$ no longer ''compresses'' this water down by some dimension(s) so that $A$ does nothing more than bijectively move this $d = \dim W$-dimensional water around.
It now makes sense to ask by how much the isomorphism $A\big\vert_W : W \to W$ is increasing or decreasing this $d$-dimensional volume, which is what $\operatorname{Tr}(A) = \operatorname{Tr}\left(A\big\vert_W\right) = \operatorname{div}\left(A\big\vert_W\right)$ represents.

Remark: This may not really answer your question since you stated that "The divergence application of trace is somewhat interesting, but again, not really what we are looking for." Nevertheless, whatever alternative non-divergence based interpretation of the trace of an isomorphism you choose, I hope that this will help you to extend it to the case where the map isn't an isomorphism.