Let us consider the basis of $U \otimes V$. As we are dealing with $\mathbb{R}^3 \otimes \mathbb{R}^3$, the basis $B$ is the set of all $e_i \otimes e_j$ for $i,j \in \{1, 2, 3\}$. Note that for the cross product, the multiplication table is as follows
$$
\begin{array}{c|c|c|c|}
\times & e_1 & e_2 & e_3 \\
\hline
e_1 & 0 & e_3& -e_2\\
\hline
e_2 & -e_3 & 0 & e_1 \\
\hline
e_3 & e_2 & -e_1 & 0 \\
\hline
\end{array}
$$
Then define a map $W: B \to \mathbb{R}^3$ by the following multiplication table
$$
\begin{array}{c|c|c|c|}
W & e_1 & e_2 & e_3 \\
\hline
e_1 & 0 & e_3& -e_2\\
\hline
e_2 & -e_3 & 0 & e_1 \\
\hline
e_3 & e_2 & -e_1 & 0 \\
\hline
\end{array}
$$
As an example of how the table is to be read, this table says that $W(e_1 \otimes e_2) = e_3$.
Now that we have defined $W$ on the basis, we can extend to the rest of $\mathbb{R}^3 \otimes \mathbb{R}^3$. This is because the tensor product and the cross product have the same structure. The cross product is a bilinear map, that is:
$$
\begin{align}
(v_1 + v_2) \times w = v_1 \times w + v_2 \times w \\
v \times (w_1 + w_2) = v \times w_1 + v \times w_2 \\
\end{align}
$$
and the tensor space follows a set of "bilinear" axioms:
$$
\begin{align}
(v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w \\
v \otimes (w_1 + w_2) = v \otimes w_1 + v \otimes w_2 \\
\end{align}
$$
These are very similar. Now, extending $W$ linearly, we truly see that
$$W(v \otimes w) = v \times w$$
As an example, which illuminates the proof of this fact, consider the following computation:
$$
\begin{align}
W((1\mathbf{e}_1 + 2\mathbf{e}_3) \otimes (3\mathbf{e}_2 + 4\mathbf{e}_3)) &= \\
W(1\mathbf{e}_1 \otimes (3\mathbf{e}_2 + 4\mathbf{e}_3) + 2\mathbf{e}_3 \otimes (3\mathbf{e}_2 + 4\mathbf{e}_3)) &= \\
W(1\mathbf{e}_1 \otimes 3\mathbf{e}_2 + 1\mathbf{e}_1 \otimes 4\mathbf{e}_3 + 2\mathbf{e}_3 \otimes 3\mathbf{e}_2 + 2\mathbf{e}_3 \otimes 4\mathbf{e}_3) &= \\
W(1\mathbf{e}_1 \otimes 3\mathbf{e}_2) + W(1\mathbf{e}_1 \otimes 4\mathbf{e}_3) + W(2\mathbf{e}_3 \otimes 3\mathbf{e}_2) + W(2\mathbf{e}_3 \otimes 4\mathbf{e}_3) &= \\
W(3(\mathbf{e}_1 \otimes \mathbf{e}_2)) + W(4(\mathbf{e}_1 \otimes \mathbf{e}_3)) + W(6(\mathbf{e}_3 \otimes \mathbf{e}_2)) + W(8(\mathbf{e}_3 \otimes \mathbf{e}_3)) &= \\
3W(\mathbf{e}_1 \otimes \mathbf{e}_2) + 4W(\mathbf{e}_1 \otimes \mathbf{e}_3) + 6W(\mathbf{e}_3 \otimes \mathbf{e}_2) + 8W(\mathbf{e}_3 \otimes \mathbf{e}_3) &= \\
3\mathbf{e}_3 - 4\mathbf{e}_2 - 6\mathbf{e}_1 + 8\cdot\mathbf{0} &= \\
-6\mathbf{e}_1 - 4\mathbf{e}_2 + 3\mathbf{e}_3
\end{align}
$$
which is indeed the cross product of those two vectors. Steps 4 and 6 rely on the fact $W$ had been extended linearly, while the rest of the steps used the bilinear properties of the tensor product. (Compare this to explicitly finding the cross product by expanding and multiplying!)
Conclusion: The tensor product allows us to use its axioms of bilinearity to "mimic" the properties of a bilinear object or map.
Let $k$ be a field. Recall that if $Z$ is a $k$-vector space, the dual space of $Z$ is the $k$-vector space $Z^*$ of all the linear maps $f : Z \to k$.
Now, if $U$ and $V$ are two $k$-vector spaces, consider the $k$-vector space $\operatorname{Bil}_k(U,V;k)$ of all the bilinear maps$^\color{blue}1$ $b : U \times V \to k$. So, here $U \otimes V$ denotes the dual space of $\operatorname{Bil}_k(U,V;k)$, that is, $$U \otimes V := \operatorname{Bil}_k(U,V;k)^*.$$ Thus, an element $z$ of $U \otimes V$ is a linear map $z : \operatorname{Bil}_k(U,V;k) \to k$. Hence, given $u \in U$ and $v \in V$, the tensor product $u \otimes v \in U \otimes V$ is the linear map $u \otimes v : \operatorname{Bil}_k(U,V;k) \to k$ such that $$\forall b \in \operatorname{Bil}_k(U,V;k) : \quad (u \otimes v)(b) = b(u,v).$$ The linearity of $u \otimes v$ follows from the definition of the operations in $\operatorname{Bil}_k(U,V;k)$: if $b_1,b_2 \in \operatorname{Bil}_k(U,V;k)$ and $\lambda \in k$, then $(\lambda b_1+b_2)(u,v) := \lambda b_1(u,v)+b_2(u,v)$.
Exercise: If $Z$ is a $k$-vector space, and $U,V$ are finite dimensional, prove that for any bilinear map $h : U \times V \to Z$ there exists a unique linear map $\tilde h : U \otimes V \to Z$ sending $u \otimes v$ to $h(u,v)$, for any $u \in U$ and $v \in V$.
$^\color{blue}1$ Which is not the same as $(U \times V)^*$.
Best Answer
No, that doesn't sound right.
First, elements of the tensor product space are not themselves bilinear maps, so they're definitely not representatives of classes of such bilinear baps.
Second $(1,1)$ is not generally something that can be the output of a bilinear map $U\otimes V\to Z$, unless $Z$ is in particular the vector space $\mathbb F^2$.
No, that can't be. If $B$ is bilinear, that requires that $B(\alpha u,v)=\alpha\cdot B(u,v)$. So if $B$ sends $(u,v)$ to $(1,1)\in\mathbb F^2$, then it must send $(\alpha u,v)$ to $(\alpha,\alpha)$ rather than to $(\alpha,1)$.
A better understanding of the tensor elements would be that $u\otimes v$ represents a recipe for applying some bilinear map you don't know what is yet. The recipe says, "once you find out what the bilinear map is, apply it to the arguments $u$ and $v$".
In this view, far from being something that represents a bilinear map, an element of $U\otimes V$ is something that will eat a bilinear map and gives you something in that map's codomain.
The reasons tensors are different from just expressions with a variable ranging over bilinear maps is that equality between tensors knows about the bilinearity. So, for example $2u\otimes v = u\otimes 2v$ because when you apply a bilinear map to either $(2u,v)$ or $(u,2v)$ you're guaranteed you get the same result.