If $E_i$ is a $\mathbb R$-vector space, then the *vector-valued trace* $\operatorname{tr}_{E_1}:(E_2\otimes E_1^\ast)\otimes(E_1\otimes E_0)\to E_1\otimes E_0$ (or *tensor contraction*) is the linearization of the (surjective) bilinear operator
\begin{equation}\begin{split}(E_2\otimes E_1^\ast)&\times(E_1\otimes E_0)\:&\to E_1\otimes E_0,\\(x_2\otimes\varphi_1)&\times(x_1\otimes x_0)&\mapsto \langle\varphi_1,x_1\rangle_{E_1}x_1\otimes x_0\end{split}\tag1\end{equation} Assume for simplicity that $E_i$ is finite-dimensional$^1$. If $E_0$ is replaced by $E_0^\ast$, then it's easy to see that $$\operatorname{tr}_{E_1}(A_2A_1)=A_2A_1\;\;\;\text{for all }A_i\in\mathcal L(E_{i-1},E_i)\tag2.$$

Is there a generalization of the kind \begin{equation}\begin{split}\bigotimes_{i=1}^kE_i\otimes E_{i-1}^\ast&\cong\mathcal L\left(\bigotimes_{i=1}^kE_{i-1},\bigotimes_{i=1}^kE_i\right)\\&\cong\bigotimes_{i=1}^k\mathcal L(E_{i-1},E_i)\ni\bigotimes_{i=1}^kA_i\mapsto A_k\cdots A_1\end{split}\tag3\end{equation} of this trace notion?

$^1$ This is clearly not necessary. We just need to replace the space $\mathcal L(E_{i-1},E_i)$ by its subset of finite rank operators.