Einstein summation convention of 2-forms: $T_{ik}\omega_{kj}$ I am familiar with Einstein summation convention that means

if the same index name appears exactly twice, once as an
upper index and once as a lower index, then that term is understood to be summed
over all possible values of that index.

But what is the meaning of the following? Let $\omega=\omega_{ij}E^i\wedge E^j$ and $T$ a (symmetric) $2$-tensor
$$T_{ik}\omega_{kj}.$$
Which index should I raise then contract it? index of $T$ or of $\omega$? in the global notation, it is $\omega(T)$ or $T(\omega)$ with appropriate converted $(1,1)$-tensors?
Also I have another problem with Einstein summation convention for 2-forms $\omega=\omega_{ij}E^i\wedge E^j$, that is the sum cannot range over all possible values of its index because $\omega=\sum_{i<j}\omega_{ij}E^i\wedge E^j$ so $i$ cannot be $n$ for example, while in $T$ it can. Unless we see $\omega$ as 2-tensors as well then the notation $\omega=\omega_{ij}E^i\wedge E^j$ is meaningless and it should be $\omega=\omega_{ij}E^i\otimes E^j$ with $\omega_{ij}=-\omega_{ji}$.
 A: I take it that you are working in a Riemannian setting so you have a metric $g_{ij}$ and its inverse $g^{ij}$ around. Then the question "which index should I raise and the contract" does not really occur, since what you want to form is $g^{k\ell}T_{ik}\omega_{\ell j}$ and by symmetry of $g^{k\ell}$ there is no choice involved. The notations $\omega(T)$ and $T(\omega)$ tend to be ambiguous anyway, in the end, the index notation is more precise here.
The two interpretations of index notation for two-forms (i.e. via $\sum_{i<j}$ or via $\sum_{i,j}$ and $\omega_{ji}=-\omega_{ij}$) are indeed there and they both are meaningful. In an interpretation involving contractions you have to take the interpretation as a tensor field with skew symmetric coefficients. Observe that also the result of your operation is a tensor field and not a two-form. I think it is also good to keep in mind that there actually is a linear algebra meaning hidden behind the index expression. You have bilinear maps $T(x),\omega(x):T_xM\times T_xM\to\mathbb R$ for which in addition you know that $\omega(x)$ is skew symmetric. Your contraction then expresses the bilinear map $(X,Y)\mapsto \sum_{i=1}^nT(X,e_i)\omega(e_i,Y)$ where $\{e_i\}$ is an orthonormal basis of $T_xM$ (and the result is independen from which orthonormal basis you uses).
