Below is a quote from this paper found on page 8.
Dually, the evaluation map is ev(T) = tr(T) for some T ∈ hom(V, V ).
Why does $ev(T)$ have to be the trace of $T$? It seems that there are a lot of maps from the vector space to the underlying field that we could use. What conditions make it necessarily the trace map?
Glad to see you're goin on studying TQFTs :)
The trace $ev$ map has a universal property; among many morphisms $V^\lor\otimes V\to k$ it's the unique map such that composed with a "dual" map $coev\colon k \to V \otimes V^\lor$ in suitable "zig-zags" $$ \begin{array}{c} V \xrightarrow{V\otimes ev} V\otimes V^\lor \otimes V \xrightarrow{coev \otimes V} V \\ V \xrightarrow{V\otimes coev} V^\lor \otimes V\otimes V^\lor \xrightarrow{ev\otimes V} V^\lor \end{array} $$ give the identity map. This definition works well in every category where there's a notion of "dual" of an object. It's a matter of a basic linear algebra course to see that if $ev$ is the trace of a linear operator and $coev$ is the map sending $1\in k$ into $\sum v_i\otimes v^i$ (subscript: elements of a basis of $V$; superscript: its dual basis) then you get precisely the zigzags identities.
There's a pretty much neat reason why this trace comes out, and that's the good monoidal structure of the category of vector spaces: the two-sided dual of $V$ is obtained as the internal hom $[V,k]$: then $ev$ and $coev$ are precisely the counit and unit, respectively, of the adjunction $-\otimes k\dashv \hom_k(-,k)$ (that's an autoequivalence of $\sf Vect$, of course), and zigzags identities are precisely those of the adjunction.
Compact closed categories arise in several places, and homming with the unit object is a general technique as well.
An engineer (no flame intended) would say that sometimes you build duals by "twisting what seems to have period 2 and hope for best"; in vector spaces you transpose a matrix. In cobordisms you invert orientation. In spectra... well, it's more complicated there but the idea is almost the same.
Since a TQFT preserves duals (that's an instructive exercise), you get the idea that $Z(S^1\coprod \overline{S^1}\to \varnothing)$ is the trace map.