What I know is that a $(q,r)$ tensor $T$ is a multilinear map $T:(V^*)^q \times V^r \to \mathbb{R}$, where $V$ is a vector space and $V^*$ is its dual space. Then it is said that a $(q,r)$ tensor can be regarded as $T : V^r \to V^q$ by thinking of the map $(v_1, ..., v_r) \to T(\cdot, v_1, ..., v_r)$ However, $T(\cdot, v_1, ..., v_r)$ is a multilinear map from $(V^*)^q$ to $\mathbb{R}$. How can I think of $T(\cdot, v_1, ..., v_r)$ as a element of $V^q$?
Also the cross product of vectors $A$ and $B$ in $\mathbb{R}^3$ is said to be a $(1,2)$ tensor. How is this possible?
Let's start slow. Lets figure out why an element of $\mathcal{T}^{1,1}(V)$ can be seen as a homomorphism from $V$ to $V$. From your definition above, $T\in\mathcal{T}^{1,1}(V)$ is a bilinear function $T:V^*\times V \rightarrow \mathbb{R}$. From this we need a recipe for sending $v\in V$ to $T(v)\in V$, and that recipe needs to be linear.
Recall that $V^{**}=V$. Hence to name an element of $V$ we need only name an element of $V^{**}$,that is a linear function that assigns to each $\phi\in V^*$ and element of $\mathbb{R}$. Wait a second $T(\, v)\in V^{**}$, and because $T$ is bilinear, when we fix the first argument we get a linear map. That's the assignment.
The operation of taking the dual commutes with the operation of taking the tensor product. So $(V\otimes V)^{*}=V^{*}\otimes V^{*}$. That means $({(V^{*})}^{\otimes r})^*=(V^{**})^{\otimes r}=V^{\otimes r}$. The full theorem is just the observation above, extended to $T\in \mathcal{T}^{r,q}(V)$.