Solution 1:

I have always viewed the term trace as used to name $$\displaystyle \operatorname{tr}(P)=\sum_{f:P\to R}f(P)$$ as a reference to the fact that that ideal is the part of $R$ which you can reach from $P$... whatever that may mean :) One can check, for example, that $P$ is a generator of the category of modules iff $\operatorname{tr}(P)=R$, and that makes sense (to me!)

A more serious connection is the following. Suppose $V$ is a finite dimensional vector space over a field $k$, and let $\hom_k(V,k)$ be its dual space. Then there is a natural isomorphism $$\phi:V\otimes\hom_k(V,k)\cong\hom(V,V).$$ On the other hand, we have the usual trace map $\operatorname{tr}:\hom(V,V)\to k$, so we can consider the composition $$\operatorname{Tr}=\operatorname{tr}\circ\phi:V\otimes\hom_k(V,k)\to k,$$ which is also sometimes called a trace map. If you now replace $k$ by ring, $V$ by a left $R$-module, then what you wrote $\operatorname{tr}(P)$ is the image of my $\operatorname{Tr}$: it follows that the trace ideal is the image of the trace map.