Given any two vector spaces $U,V$ over the same field $\Bbb{F}$, we can define their tensor product $U\otimes V$ via the universal property you mention in the first paragraph. In the finite-dimensional case, we have the following canonical isomorphism:
\begin{align}
U\otimes V \cong (U^{**})\otimes (V^{**})\cong (U^*\otimes V^*)^*\cong \text{Hom}^2(U^*\times V^*;\Bbb{F}),
\end{align}
where this last space is the space of bilinear maps $U^*\times V^*\to \Bbb{F}$, and the last isomorphism is literally by the universal property (replace each of $U,V, W$ respectively with $U^*,V^*,\Bbb{F}$ in your first diagram). The full isomorphism is such that an element $u\otimes v$ of the LHS is identified with the bilinear map $(\xi,\eta)\mapsto \xi(u)\cdot \eta(v)$.
Now, if you fix a single vector space $V$ over a field $\Bbb{F}$, and integers $r,s\geq 0$, then we define the $(r,s)$-tensor space over $V$ to be
\begin{align}
T^r_s(V)&:= \underbrace{V\otimes \cdots \otimes V}_{\text{$r$ times}}
\otimes \underbrace{V^*\otimes \cdots \otimes V^*}_{\text{$s$ times}}.
\end{align}
By similar reasoning as above, if $V$ is finite-dimensional we see that (invoke double duality as needed):
\begin{align}
V^{\otimes r}\otimes (V^*)^{\otimes s} &\cong \text{Hom}^{r+s}\bigg((V^*)^r\times V^s\,\,;\,\,\Bbb{F}\bigg),
\end{align}
and the isomorphism is such that $v_1\otimes \cdots \otimes v_r\otimes \alpha^1\otimes \cdots \otimes\alpha^s$ on the LHS is identified with the multilinear map
\begin{align}
(\beta^1,\dots, \beta^r, w_1,\dots, w_s)\mapsto \beta^1(v_1)\cdots \beta^r(v_r)\cdot\alpha^1(w_1)\cdots \alpha^s(w_s).
\end{align}
Hopefully this clears up the issue of what the isomorphism are (I leave it to you to verify these really are isomorphisms). For your final question of whether tensors are multilinear maps, well, I feel like now we're getting into a terminology issue. To me a tensor is an element of a tensor product of two vector spaces. What is a tensor product? From the universal property, we know such a construction is unique up to a unique isomorphism (as always). Therefore, the question of what a tensor is, is rather unimportant I feel.
If you use the realization of a tensor product as a quotient of some free module, then your tensors are elements of that space. If on the other hand you're dealing with finite-dimensional spaces, then you can equivalently define tensors as certain multilinear maps, so to you a tensor could be a multilinear map.
My Introduction to Tensors and Opinion:
The first definition of tensor I encountered was the one with multilinear maps (as introduced in Spivak), and I found this straight-forward, and I'm glad this is the first definition I learnt, because I don't think I would have been able to digest the abstractness of the universal property definition. Also, this definition is rather "concrete", and very easy to state given only basic knowledge of linear algebra (i.e just knowing what a vector space, dual space and multilinear means) so it made it easy for me to see what tensors are and work with them very easily, even though that's simply not that relevant (it's just one of those things which gives one some comfort when first learning a new topic). This concrete realization of tensors meant I didn't have to take a long abstract-algebraic detour in the middle of my advanced calculus and physics course into universal properties, quotient vector spaces and so on. This context hopefully puts it into some perspective for why people introduce tensors on finite-dimensional spaces as multilinear maps, and hence say that "tensors are multilinear maps".
Also, on a real-vector space, such as $\Bbb{R}^n$, two very common and important examples of tensors are the inner product (a $(0,2)$ tensor) and the determinant/volume form (a $(0,n)$ tensor). These are already conveniently given to us in the form of multilinear maps. So on top of wanting to avoid excessive algebraic detours, the fact that very important tensors (and by extension, tensor fields on manifolds) are already presented in multilinear fashion further motivates the multilinear definition one usually encounters in differential geometry/physics, hence the frequently encountered statement "tensors are multilinear maps".
Best Answer
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.