Does the index notation of tensor require the same set of basis for each copy of V? According to Wikipedia, when defining tensor using tensor product, each copy of V(or V*) seems to use the same set of basis (or co-basis), but I feel it is ok to use different irrelevant basis for each copy of V and V*. More rigorous description of the question:
$$ T\in V\otimes\dotsb\otimes V\otimes V^*\otimes\dotsb\otimes V^*$$
$$T = T^{i_1\dotsm i_p}_{j_1 \dotsm j_q} e^1_{i_1} \otimes \dotsb \otimes e^p_{i_p} \otimes \epsilon_1^{j_1} \otimes \dotsb \otimes \epsilon_q^{j_q} .$$
here $e^m_{i_m}$ is from the choosen set of basis of the mth copy of $V$, $\epsilon_n^{j_n}$ is from the choosen set of basis of the nth copy of $V^*$. These choosen basis from different copy of $V$ and $V^*$ are not related at all. (For comparison, the definition in Wikipedia seems to require all the choosen basis of $V$ are the same, and all the choosen basis of $V^*$ are its dual basis, the question is about removing this requirement)
Then does the above $T^{i_1\dotsm i_p}_{j_1 \dotsm j_q}$ still transform like a tensor when a given linear transform (and its dual linear transform) is applyied to the choosen basis of each V (and V*)?
 A: One of the most important properties of tensors is that the set of all tensors of type $(r,s)$ at a point $P$ constitutes a vector space of dimension $n^{r+s}$. It is so because you can add two tensors at P and get a new tensor of the same type $(r,s)$ and you can multiply a tensor by a scalar or a constant and still yield a tensor of the same type.
With your proposed system this is no longer possible because you are not allowed to add tensors whose types are distinct. You need to complicate matters by introducing new types and add some transformations.
We would also like to be able to multiply the components of two tensors of types $(r_1,s_1)$ and $(r_2,s_2)$ to yield a tensor of type $(r_1+r_2, s_1+s_2)$. Again this will not be possible with your proposal without a profusion of new definitions and transformations.
Another important property of tensors is the invariance under contraction. We need stuff like
$$
\mathbf{a}\cdot(\nabla\times V)=\varepsilon^{ijk}\frac{\partial V_j}{\partial x^i}a_k
$$
to be invariant regardless of the underlying charts. Imagine the work needed to be put in if you expressed the $(0,2)$-tensor $V_{j|k}=\frac{\partial V_j}{\partial x^k}-\Gamma_{j\,\,\,k}^{\,\,m}V_m$ in random charts for each component.
Not to mention the profusion of complications dealing with differential forms and so on.
