Tensor contraction (vector-valued trace) on $\bigotimes_{i=1}^k\mathcal L(E_{i-1},E_i)$

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 $$$$\begin{split}(E_2\otimes E_1^\ast)&\times(E_1\otimes E_0)\:&\to E_1\otimes E_0,\\(x_2\otimes\varphi_1)&\times(x_1\otimes x_0)&\mapsto \langle\varphi_1,x_1\rangle_{E_1}x_1\otimes x_0\end{split}\tag1$$$$ Assume for simplicity that $$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 $$$$\begin{split}\bigotimes_{i=1}^kE_i\otimes E_{i-1}^\ast&\cong\mathcal L\left(\bigotimes_{i=1}^kE_{i-1},\bigotimes_{i=1}^kE_i\right)\\&\cong\bigotimes_{i=1}^k\mathcal L(E_{i-1},E_i)\ni\bigotimes_{i=1}^kA_i\mapsto A_k\cdots A_1\end{split}\tag3$$$$ 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.