What allows a bilinear form (output: field element) to also work as a linear map (output: vector)? 
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*A bilinear form $B: V × V → K$, when the inputs are 2 vectors, has 1 element of their field as output.

*A linear map $L: V → W$, when the input is 1 vector, has 1 vector as output.

I have seen cases of maps that worked as bilinear forms and simultaneously as linear maps, e.g. the metric tensor of rank (0,2); consider the expression $g_{\mu\nu}dx^\mu dx^\nu$:

*

*As a bilinear form, it takes two vectors as inputs and it outputs 1 field element (e.g. if the inputs are the local position vectors, it outputs the value of the line element);

*As a linear map, it takes only the first vector as an input, and it outputs its dual:  $g_{\mu\nu}dx^\mu dx^\nu = dx_\nu dx^\nu$. In this way, the metric tensor working as a linear map is acting as the "dual converter".

Note that I am not interested in the reason why the results come out the same in both cases (it's trivial to understand). I am interested in the mechanism and conditions that allow a map to have such double-face.
I would like to know: is it always the case that a bilinear form (2 inputs, 1 output) can be interpreted also as a linear map (1 input, 1 output)? If not, what requirements are there for a bilinear form (2 inputs, 1 output), to work also as a linear map (1 input, 1 output)?
 A: A bilinear map $B\colon U\times V \to W$ gives for every $u\in U$ a linear map $B(u,-)\colon V \to W$. This assignment can be seen as a map $U\to L(V;W)$, $u\mapsto B(u,-)$, which is easily verified to be linear. Similarly, it also gives a linear map $V\to L(U;W)$, $v\mapsto B(-,v)$. So except for linearity there are no additional conditions.
Note that $L(V;K) = V^*$ per definition, so for bilinear forms we indeed get maps to the duals $V^*$ and $U^*$.
A: You can view bilinear forms as elements of $V^* \otimes V^*$, as you pointed out. Explicitly, if you have a basis $e_i$ with dual basis $e^i$ then
$$
B = B(e_i, e_j)e^i \otimes e^j
$$
and so
$$
\text{Bilinear forms(V)} \cong V^* \otimes V^*
$$
In the same way, you can view $V^* \otimes W$ as maps from $V$ into $W$. If $T : V \to W$ then we can write
$$
T(e_i) = T_i^jf_j
$$
where $f_j$ is a basis of $W$. Hence
$$
T = T_i^j e^i \otimes f_j
$$
and so
$$
\text{Hom}(V, W) \cong V^* \otimes W
$$
in a natural way. So what is happening in your example is that you can either think of the first isomorphism or the second isomorphism as $W = V^*$
