Suppose $k$ is a commutative ring, and $V$ is a $k$-module. Let $\operatorname{End}_V$ be the functor $\mathsf{Alg}_k\to\mathsf{Set}$ defined by $\operatorname{End}_V(R)=\operatorname{End}_{R-lin}(R\otimes_k V)$. If $V$ is finitely generated projective, why is $\operatorname{End}_V$ representable by $\operatorname{Sym}(V\otimes_k V^\vee)$?
This is stated in the first half of Example 4.1 of JS Milne's Algebraic Groups, in showing that this functor is an algebraic monoid.
The claim is that there is a natural bijection $\text{End}_R(R \otimes V) = \text{Hom}_{k-\text{Alg}}(\text{Sym}(V \otimes V^\vee), R)$. By the universal property of the symmetric algebra, the latter is $\text{Hom}_{k-\text{mod}}(V \otimes V^\vee, R)$.
There is a map $ \text{End}_R(R \otimes V)\to \text{Hom}_{k-\text{mod}}(V \otimes V^\vee, R) $ given by the rule
$$M \mapsto [v \otimes f \mapsto f(Mv)].$$
This map is an isomorphism when $V$ is finitely generated free, because it is injective and both sides are $R$-modules of the same rank. Since you may check that this map is an isomorphism Zariski locally on $R$, it follows that it is also an isomorphism when $V$ is finitely generated projective. (I am not positive about this last argument.)