$\def\vc#1{\vec{\mathbf{#1}}}$
$\def\cv#1{\tilde{\mathbf{#1}}}$
$\def\qty#1{\left(#1\right)}$
$\def\F{\mathbb{F}}$
In a number of standard textbooks on tensors for physics students (e.g. Tensors: The Mathematics of Relativity Theory and Continuum Mechanics) we find definitions of the tensor product like
$$
[ \cv{u}\otimes \cv{v}]\qty{\vc{a},\vc{b}}:=\cv{u}(\vc{a})\cv{v}(\vc{b}) \tag{I}
$$
where $\cv{u},\cv{v}$ are covectors and $\vc{a},\vc{b}$ are vectors. Reading this expression, $\cv{u}\otimes \cv{v}$ seems to be a bilinear form $V \times V \rightarrow \F$ ($V$ is the vector space over a field $\F$).
However the universal property of the tensor product only allows for any bilinear map
$B: V^* \times V^* \rightarrow \F$ to be factored through a unique linear map $\beta: V^* \otimes V^* \rightarrow \F$ so that
$$
\beta(\cv{u} \otimes \cv{v}) = B(\cv{u},\cv{v})
$$
Therefore bilinear forms can only be linked to a tensor product through a linear map.
Based on this can we say that the tensor product of two covectors is a bilinear form?
Best Answer
Tensor products
A canonical definition of a tensor product is by the universal property:
The tensor product of the vector spaces $U$, $V$ is a vector space $U \otimes V$, together with a bilinear function $\otimes : U \times V \to U \otimes V$, satisfying the property
The definition is correct because we can prove that given $U$ and $V$, there is a unique space up to isomorphism satisfying the universal property.
Relation to bilinear forms
In the case of tensor products of the form $V^* \otimes V^*$, its elements can be canonically interpreted as bilinear forms $V \times V \to \mathbb{F}$. Formally there is a canonical vector space monomorphism $j : V^* \otimes V^* \to \operatorname{Bil}(V \times V, \mathbb{F})$, satisfying
$$j(u \otimes v)(a, b) = u(a) \cdot v(b).$$
If $V$ is finitely dimensional, $j$ is surjective and so an isomorphism. Then we can freely identify every tensor with its corresponding bilinear forms, so that simple tensors $u \otimes v$ are mapped to the forms $B(a, b) = u(a) \cdot v(b)$, and arbitrary tensors (finite sums of simple tensors) are mapped adequately by additivity. With $V$ infinitely dimensional the identification is still possible, except not all bilinear forms will be expressible as a tensor.
Conclusion
The definition of a tensor product via the universal property is powerful and convenient, but takes some understanding to master. On the other hand, bilinear forms are quite easy to grasp and in specific cases they are equivalent to tensors. My guess is that physicists want to avoid the complications of the general tensor product and so define tensors directly as bilinear forms, which is sufficiently true for their purposes.