How can you do the tensor product between the basis covectors $ \epsilon^i $ & $ \epsilon^j $ to result in $ \epsilon^i \otimes \epsilon^j $ when $ \epsilon^i $ & $ \epsilon^j $ have only contravariant indices?
I did check this- How is it that two covariant vectors can be multiplied without one being transposed?
But there it was assumed $ \epsilon^i \otimes \epsilon^j $ to be true by default.
If $\varepsilon^i$ are linear functionals (a.k.a covectors), that is, $\varepsilon^i:V\to\mathbb R$ are linear transformations, then the $\varepsilon^i\otimes\varepsilon^j$ are bilinear transformation $$\varepsilon^i\otimes\varepsilon^j:V\times V\to\mathbb R$$ defined by $$\varepsilon^i\otimes\varepsilon^j(v,w)=\varepsilon^i(v)\varepsilon^j(w).$$ Linearity of $\varepsilon^i$ implies the bi-linearity of the $\varepsilon^i\otimes\varepsilon^j$. In general $\varepsilon^i\otimes\varepsilon^j$ is different from $\varepsilon^j\otimes\varepsilon^i$ and the $\varepsilon^j\otimes\varepsilon^j$ make sense.