How can linear transformations and the cross product be respresented as a $(1, 1)$ tensor and a $(1, 2)$ tensor, respectively? I understand how $(0, 1)$ and $(1, 0)$ tensors are linear functional and vectors, respectively. But I can't see how $(1, 1)$ tensor can represent linear transformations, and how $(1, 2)$ tensors can represent the cross product. I have shown below how I 'derived' that linear functionals and vectors are $(0, 1)$ and $(1, 0)$ tensors. How can I do the same for linear transformations and the cross product?
(0, 1) Tensor
\begin{array}{rll}
T: & V & \longrightarrow \mathbb{R} \\
& v & \longmapsto \psi(v) 
\end{array}
So $\psi \in V^\ast$, that is, $\psi$ is a linear functional.
(1, 0) Tensor
\begin{array}{rll}
T: & V^\ast & \longrightarrow \mathbb{R} \\
& \psi & \longmapsto v(\psi) = \psi(v)
\end{array}
So $v \in V$, that is, $v$ is a vector.
(1, 1) Tensor
\begin{array}{rll}
T: & V^\ast \times V & \longrightarrow \mathbb{R} \\
& (\psi, v) & \longmapsto  ???
\end{array}
???, is a linear transformation.
(1, 2) Tensor
\begin{array}{rll}
T: & V^\ast \times V \times V& \longrightarrow \mathbb{R} \\
& (\psi, v, w) & \longmapsto  ???
\end{array}
???, is a cross product.
 A: For the $(1,1)$ tensor: if $\Phi:V \to V$ is a linear transformation, then we can take
$$
T: (\psi,v) \mapsto \psi(\Phi(v))
$$
I would not say that every $(1,2)$ tensor is a cross-product, but the cross product is an example of such a tensor.  In particular, we have the map
$$
T: (\psi,v,w) \mapsto \psi(v \times w)
$$
which satisfies the requirements of a $(1,2)$ tensor.
A: For vector spaces $V_{1},\dots,V_{r}$ and $W$ (over any field $\mathbb{K}$), you have a natural isomorfphism
\begin{align*}
L_{\mathbb{K}}^{k}(V_{r-k},\dots,V_{r};L_{\mathbb{K}}^{r-k-1}(V_{1},\dots,V_{r-k-1};W)) \cong L_{\mathbb{K}}^{r}(V_{1},\dots,V_{r};W)
\end{align*}
associanting the map $G: V_{r-k} \times \cdots \times V_{r} \to L_{\mathbb{K}}^{r-k-1}(V_{1},\dots,V_{r-k-1};W)$ to the map $T_{G} \in L_{\mathbb{K}}^{r}(V_{1},\dots,V_{r};W)$ defined by
\begin{align*}
T_{G}(v_{1},\dots,v_{r})=G(v_{r-k},\dots,v_{r})(v_{1},\dots,v_{r-k-1}).
\end{align*}
Now for finite dimensional vector spaces $V_{1},\dots,V_{r}$, taking the construction of the tensor product as multilinear maps, i.e., $V_{1} \otimes \cdots \otimes V_{r}:=L_{\mathbb{K}}^{r}(V_{1}^{*},\dots,V_{r}^{*};\mathbb{K})$ with $v_{1} \otimes \cdots \otimes v_{r}$ being the map $(\psi^{1},\dots,\psi^{r}) \mapsto \psi^{1}(v_{1}) \cdots \psi^{r}(v_{r})$, if you apply this to 
\begin{align*}
V^{\otimes_{r}} \otimes (V^{*})^{\otimes_{s}}=L_{\mathbb{K}}^{r+s}(\underbrace{V^{*},\dots,V^{*}}_{\text{$r$ times}},\underbrace{V,\dots,V}_{\text{$s$ times}};\mathbb{K}),
\end{align*}
where I put $V^{**}=V$ since everything is finite dimensional, the case of $(1,1)$ tensors is given by the isomorphisms
\begin{align*}
V \otimes V^{*} \cong L_{\mathbb{K}}(V;L(V^{*};\mathbb{K})) \cong L_{\mathbb{K}}(V,V)
\end{align*}
where you take the decomposable tensors $v \otimes \psi$ and identify it with the linear map $T_{v,\psi} : w \mapsto \psi(w)v$, and extending it linearly to the whole space.
Now take $\dim V=3$ and $\mathbb{K}=\mathbb{R}$. To have a cross product you need an inner product $\langle \cdot, \cdot \rangle$ on $V$. The cross product $\times : V \times V \to V$ associated with this inner product is a member of $L^{2}(V,V;V)$ and we have the isomorphisms
\begin{align*}
L^{2}(V,V;V) \cong L^{2}(V,V;L(V^{*};\mathbb{R})) \cong L(V;L^{2}(V^{*},V;\mathbb{R})) \cong L^{3}(V^{*},V,V;\mathbb{R})=V \otimes V^{*} \otimes V^{*}
\end{align*}
For a decomposable $(1,2)$ tensor $v \otimes \psi \otimes \varphi$ you can have the $T \in L^{2}(V,V;V)$ map given by $T(u,w)=\psi(u) \varphi(w) v$, being clear what to associate with and non-decomposable tensor. For the cross product you can have the tensor
\begin{align*}
\sum_{i,j,k=1}^{3} \varepsilon_{ijk} \, e_{k} \otimes \epsilon^{i} \otimes \epsilon^{j}
\end{align*}
where $\{e_{1},e_{2},e_{3}\}$ is an orthonormal base such that $e_{1} \times e_{2} = e_{3}$ and $\{\epsilon^{1},\epsilon^{2},\epsilon^{3}\}$ is its dual base.
You can clearly see that the map associated with it is the cross product: if $T_{k}^{\, ij}$ is the map associated with $e_{k} \otimes \epsilon^{i} \otimes \epsilon^{j}$ then
\begin{align*}
\left( \sum_{i,j,k=1}^{3} \varepsilon_{ijk} \, T_{k}^{\, ij} \right) (v,w) &= \sum_{i,j,k=1}^{3} \varepsilon_{ijk} \, \epsilon^{i}(v) \, \epsilon^{j}(w) \, e_{k}\\
&= v \times w
\end{align*}
