I have an intuition, but not sure exactly, whether the partial trace is additive in the sense of direct sum. My intuition is that partial trace is additive and direct sum acts as a sum but with orthogonality (I hope I understand that right).
I mean: $Tr_{E_1} V_1 \oplus Tr_{E_2}V_2 = Tr_{E_1 \oplus E_2} (V_1 \oplus V_2) $.
Actually this kind of means that $(E_1 \otimes B_2) \oplus (E_2 \otimes B_2) = (E_1 \oplus E_2)\otimes(B_1\oplus B_2)$. Which I am not sure, but might be legit.
Is that true? Assuming $V_i: B_i \otimes E_i \longrightarrow B_i \otimes E_i$.
If $B_1 = B_2 = B$, then it is true in a sense, but you have to be careful about what you mean by $\operatorname{Tr}_{E_1 \oplus E_2}$. In particular, we note that the spaces $$ (E_1 \otimes B) \oplus (E_2 \otimes B), \quad (E_1 \oplus E_2) \otimes B $$ are canonically isomorphic. In particular, the linear map $\Phi: (E_1 \oplus E_2) \otimes B \to$ $(E_1 \otimes B) \oplus (E_2 \otimes B)$ defined such that $ \Phi[(x_1 \oplus x_2) \otimes y] = (x_1 \otimes y) \oplus (x_2 \otimes y) $ defines an isomorphism.
With that said, your statement can be rendered as $$ \operatorname{tr}_{E_1 \oplus E_2}(\Phi^{-1} \circ (V_1 \oplus V_2) \circ \Phi)= \operatorname{tr}_{E_1}(V_1) \oplus \operatorname{tr}_{E_2}(V_2). $$