How does a 4-tensor linearly trasform an arbitrary 2-tensor? I'm trying to understand tensors by looking at this table and thinking about the various types of transformations the tensors represent. From the linked table, I tried looking up some of the less familiar examples of tensors, such as the elasticity tensor, to understand things. 
The linked page for elasticity tensors on Wikipedia says "...here c is a fourth-order tensor (that is, a linear map between second-order tensors)", without much other context. If a (1,1) tensor is a linear transformation, a (2,2) tensor being a linear transformation of 2-tensors makes sense (even though the article says 4-tensor), but how does a (2,2) tensor act on arbitrary 2-tensors? (which I assume means any kind of 2-tensor -- linear transformations, bilinear forms, and bi-vectors, all of which I am familiar with) How is the computation actually preformed for the various types of 2-tensors? I assume it is in terms of operations like contraction and the tensor product, but I just can't seem to find a reasonable explanation. 
 A: This is a really old question but I just recently had to go through some of this stuff so ended up stumbling across it while searching.
OK so there are various basic properties about the tensor product which are useful for getting off the ground:
Given any finite dimensional vector space $V$, we have that $V \otimes V^* \simeq \mathrm{End}(V),$ the space of linear maps from $V$ to itself.
Apply the above fact to the vector space $V \otimes V^*$ and you get that $$(V \otimes V^*) \otimes (V \otimes V^*)^* \simeq \mathrm{End}(V \otimes V^*).$$
On the left hand side we can do some simplification because

*

*The dual space of a tensor product is the tensor product of the dual spaces; and

*The tensor product of vector spaces is associative and commutative,

So $(V \otimes V^*) \otimes (V \otimes V^*)^*$ is 'unambiguously' (isomorphic to) $V\otimes V \otimes V^* \otimes V^*$, the space of $(2,2)$ tensors. But on the other hand, using the same properties, it is clearly equal to $\mathrm{End}(V) \otimes \mathrm{End}(V)$. And on the right-hand side, we notice that $\mathrm{End}(V \otimes V^*) \simeq \mathrm{End}(\mathrm{End}(V))$, i.e. we have the space of linear maps from matrices to matrices.

But you're asking more about how they act. There basically isn't just one answer; one of the things about all this algebraic stuff is that there are many ways of looking at the same thing and you need to pick a frame that suits what you are actually doing with these objects.
To interpret how a tensor in $\mathrm{End}(V) \otimes \mathrm{End}(V)$ acts as a linear map from $\mathrm{End}(V)$ to $\mathrm{End}(V)$, look at how elementary tensors act. Everything is a square matrix over $V$, i.e. we have $A \otimes B$ and it should act as a linear map on the matrix $M$. To cut a long story short its $M \mapsto AMB^T$.
This again goes via various unravellings of the definitions until you can 'see' it... think of $\mathrm{End}(V)$ as $V \otimes V^*$ again. So then we have an elementary element of $$(V\otimes V^*) \otimes (V^* \otimes V) \simeq \mathrm{End}(V) \otimes \mathrm{End}(V^*)$$ acting on something in $\mathrm{End}(V) \simeq V \otimes V^*$. And the answer to this question will also help - essentially when we use matrices, the rows are covectors and the columns are vectors, so to interpret something like
$$(A \otimes B)(x\otimes y) = A(x) \otimes B(y)$$
in actual matrices, what's happening is that $B$ is acting on the columns of the matrix that $x\otimes y$ represents. And a matrix acting on columns is given by right-multiplying by the transpose.
