Given a finite dimensional vector space $V$, how do elements of $\text{Sym}^2(V)$ give a map $V^*\rightarrow V$? An article I'm reading claims that if $V$ is a finite dimensional vector space over a field $k$, and $f\in \text{Sym}^2(V)$, then $f$ induces a morphism $V^*\rightarrow V$, where $V^*$ denotes the dual space.
If we let $x_1,\ldots,x_r$ be a basis of $V$, then we can write $f = \sum_{i\le j}a_{ij}x_ix_j$, and given $\varphi\in V^*$, one could try to consider the map
$$\varphi\mapsto \sum_{i\le j}a_{ij}\varphi(x_i)x_j\in V$$
but surely this isn't well defined...
EDIT: This is the bottom of page 38 of the stacks project.
 A: There is a tensor-hom adjunction $\hom(V,W)\cong V^\vee\otimes W$, where $\varphi\otimes w$ acts as a linear map $V\to W$ via the relation $(\varphi\otimes w)(v)=\varphi(v)w$.
Combining this with the natural isomorphism $V\to V^{\vee\vee}$ (for finite-dimensional $V$, otherwise it is simply a natural transformation), we have
$$ V^{\otimes 2}=V\otimes V\cong V^{\vee\vee}\otimes V\cong \hom(V^{\vee},V).$$
Thus, any tensor in $V^{\otimes 2}$ may be regarded as a linear map $V^{\vee}\to V$. Here's how: let $v_1\otimes v_2$ be an element of $V\otimes V$, then define $V^{\vee}\to V$ by $(v_1\otimes v_2)\varphi=\varphi(v_1)v_2$.
In particular, we may regard $\mathrm{Sym}^2(V)$ as the $S_2$-invariant subspace of $V^{\otimes 2}$, and so any element of $\mathrm{Sym}^2(V)$ induces a linear map $V^{\vee}\to V$.
If you regard $\mathrm{Sym}^2(V)$ as a quotient of $V^{\otimes 2}$ with elements written as sums of products $v_1v_2$, then the subspace and the quotient definitions are isomorphic via the symmetrization
$$ v_1\otimes v_2+v_2\otimes v_1 ~ \longleftrightarrow ~~ v_1v_2 $$
(up to scaling), in which case if you want to describe how to use elements of $\mathrm{Sym}^2(V)$ as linear maps $V^{\vee}\to V$ with the quotient definition, we can simply write
$$ (v_1v_2)\varphi:=(v_1\otimes v_2+v_2\otimes v_1)\varphi=\varphi(v_1)v_2+\varphi(v_2)v_1. $$
A: Define $f:V\otimes_KV\to\mathrm{Hom}_K(V^*,V)$ by
$$
  f(v\otimes w)(\phi)=\phi(v)w+\phi(w)v.
$$
This is clearly symmetric under $v\otimes w\mapsto w\otimes v$, so it factors to a map $\mathrm{Sym}^2(V)\to\mathrm{Hom}_K(V^*,V)$.
As Lord Shark the Unknown comments, the source link gives a matrix for the transformation using a basis for $V$. This can be verified to agree with the above.
