The matrix of the bilinear form $g = \epsilon ^1 \otimes \epsilon ^2 + 2 \epsilon ^2 \otimes \epsilon ^2$ If I have some simple bilinear form (tensor) example given as:
$g = \epsilon ^1 \otimes \epsilon ^2 + 2 \epsilon ^2 \otimes \epsilon ^2$
The epsilons denote vectors (covectors) of the dual basis $K^*$.
The matrix of the bilinear form is: 
$$[g]_K = \begin{pmatrix} 0 & 1 \\ 0 & 2 \end{pmatrix}$$
But the coefficients of the matrix $[g]_K$ have actually lower indices, like:
$$[g]_K = \begin{pmatrix} g_{11} & g_{12} \\ g_{21} & g_{22} \end{pmatrix}$$
Do I understand it correctly?
 A: The indices in the coefficients come from the objects where you evaluate the tensor on. Since $g$ is covariant, it eats vectors. If $(e_1,e_2)$ are dual to $(\epsilon^1, \epsilon^2)$, you can evaluate $g_{ij} = g(e_i,e_j)$, but not $g^{ij} = g(\epsilon^1,\epsilon^2)$, as the latter a priori does not make sense. Then $$\begin{align} g_{11} &= g(e_1,e_1) = (\epsilon^1\otimes \epsilon^2)(e_1,e_1) + 2(\epsilon^2\otimes \epsilon^2)(e_1,e_1) \\ &=  \epsilon^1(e_1)\epsilon^2(e_2) + 2\epsilon^2(e_1)\epsilon^2(e_1) \\ &= 1\cdot 0 + 2 \cdot 0 \cdot  0 \\ &= 0,\end{align}$$and similarly for the others.
If instead you had something contravariant like $h = e_1\otimes e_1\otimes e_2$, it would make sense (under the natural identification between a finite-dimensional vector space and its bidual) to evaluate $h$ in covectors, so that, for example, $$h^{112} = h(\epsilon^1,\epsilon^1,\epsilon^2) = (e_1\otimes e_1\otimes e_2)(\epsilon^1,\epsilon^1,\epsilon^2) = \epsilon^1(e_1)\epsilon^1(e_1)\epsilon^2(e_2) = 1\cdot 1 \cdot 1 = 1,$$while $h^{122} = 0$, etc..
