I know via isomorphism we may treat the tangent space of a point on a manifold as the vector space of derivations on functions at that point. I.e. we can give the tangent space the basis of partials: $$ v=v^i \partial_i $$
This allows vectors to act on functions, and also on eachother via the Lie bracket.
My question is, for tensor products of the tangent space, can we still act on function or vectors?
For a 2-tensor $T$: $$ T= T^{ij} \partial_i \otimes \partial_j $$
Can $T$ be made to act on functions or vectors in an unambiguous way?
In my opinion, the conceptual way to understand this is by viewing the action of a vector field $\xi$ on a function $f$ as the pairing between $\xi$ and the one-form $df$. In this language this becomes a point-wise question and, multilinear algebra tells you, what you can do canonically with a tensor field. Taking the example of $T=T^{ij}\partial_i\otimes\partial_j$ that you mention, there are several possibilities. As mentioned in the comment of @ziggurism , you can take smooth functions $f$ and $g$ and then form $T(df,dg)=\sum_{i,j} T^{ij}\partial_i(f)\partial_j(g)$. This corresponds to the interpretation of the values of $T$ as a bilinear formson the cotangent spaces. Alternatively, you can view $T$ as a map sending covectors to vectors, so you can associate to $df$ the vector field $T(df,\_)$ which is given by $\sum_j b^j\partial_j$, where $b^j=\sum_iT^{ij}\partial_j(f)$. There are more possibilities in the presence of a Riemannian metric, that you can use to raise and lower indices.