I quickly reviewed the following article of David Rydh.

David Rydh.

Families of Cycles.

2008.

https://people.kth.se/~dary/famofcycles20080518.pdf

Rydh extends to positive characteristic the definition of Angeniol in characteristic $0$. I will explain the construction in your special case. The key algebra lemma has to do with the $n\times n$ determinant. For a unital, commutative ring $R$, for a free $R$-module $F$ of rank $n$, for the associated free $R$-module $E=\text{Hom}_R(F,F)$ of rank $n^2$ with its structure of associative, unital $R$-algebra, the usual determinant,
$$
\text{det}_{F}:E \to R,
$$
comes from an element
$$
\text{det}'_F\in \text{Sym}_R^n(E^\vee),
$$
where for every $R$-module $M$, $M^\vee$ is shorthand for $\text{Hom}_R(M,R)$. Assuming that $n!$ is invertible in $R$, this element is equivalent to a homomorphism of $R$-modules,
$$
\widetilde{\text{det}}_F:(E^{\otimes n})^{\mathfrak{S}_n} \to R,
$$
where for an $R$-module $M$, $M^{\otimes n}$ denotes the $n$-fold tensor product $M\otimes_R \dots \otimes_R M$, and where for an $R[\mathfrak{S}_n]$-module $N$, $N^{\mathfrak{S}_n}$ denotes the $R$-submodule of $\mathfrak{S}_n$-invariants. The determinant homomorphism is normalized so that for an ordered basis $(\mathbf{e}_1,\dots,\mathbf{e}_n)$ of $F$ and for the associated basis $(\mathbf{e}_{i,j})_{1\leq i,j\leq n}$ of $E$, this $R$-module homomorphism sends the $\mathfrak{S}_n$-invariant, idempotent element $$\epsilon_{\text{id}} := \sum_{\sigma \in \mathfrak{S}_n} \mathbf{e}_{\sigma(1),\sigma(1)}\otimes \dots \otimes \mathbf{e}_{\sigma(n),\sigma(n)}$$ to $1$.

The algebra fact is that for a commutative $R$-subalgebra $A$ of $E$, the restriction of $\widetilde{\text{det}}_F$ to $(A^{\otimes n})^{\mathfrak{S}_n}$ is a homomorphism of commutative $R$-algebras. For a choice of ordered basis as above, for the subalgebra $A$ spanned by $\mathbf{e}_{1,1},\dots,\mathbf{e}_{n,n}$, this can be checked directly. The $R$-algebra $(A^{\otimes n})^{\mathfrak{S}_n}$ is isomorphic to a direct product of factors of $R$ indexed by the nondecreasing functions $u:\{1,\dots,n\}\to \{1,\dots,n\}$ (with the standard total order). The corresponding idempotent is $$\epsilon_u := \frac{1}{\# \text{Aut}(u)}\sum_{\sigma\in \mathfrak{S}_n} \mathbf{e}_{u(\sigma(1)),u(\sigma(1))}\otimes \dots \otimes\mathbf{e}_{u(\sigma(n)),u(\sigma(n))}.$$ The idempotent $\epsilon_{\text{id}}$ corresponds to the identity function on $\{1,\dots,n\}$. The determinant $R$-module homomorphism sends the idempotent $\epsilon_{\text{id}}$ to $1$, and it sends every other idempotent to $0$.

Returning to the setup in the question, denote by $\text{pr}_S:X\times S \to S$ the projection morphism. By hypothesis, the $\mathcal{O}_S$-module $F:=\text{pr}_{S,*}\mathcal{F}$ is a locally free $\mathcal{O}_S$-module of rank $n$. For every open affine subscheme $U\subset X$, there exists a maximal open subscheme $V\subset S$ such that $\text{Supp}(\mathcal{F}) \cap (V\times X)$ is contained in $V\times U$. Since $X$ is projective, as $(U,V)$ varies, the open subsets $V$ form an open cover of $S$. Thus, it suffices to construct the "Hilbert-Chow morphisms" on the open subsets $V$, provided those local morphisms satisfy the glueing condition on overlaps.

For every element $a\in \Gamma(U,\mathcal{O}_U)$, left multiplication by $a$ on $\mathcal{F}|_{V\times U}$ defines an element $\widetilde{a}$ in the endomorphism algebra, $$E := \text{Hom}_{\mathcal{O}_V}(F|_V, F|_V).$$ This defines a homomorphism of $\mathcal{O}_V$-algebras of quasi-coherent sheaves,
$$ \phi : \Gamma(U,\mathcal{O}_U)\otimes_{\mathbb{C}} \mathcal{O}_V \to E.$$ Taking tensor products, this defines a homomorphism of $n$-fold tensor product $\mathcal{O}_V$-algebras, $$ \phi^{\otimes n}: \Gamma(U,\mathcal{O}_U)\otimes_{\mathbb{C}} \dots \otimes_{\mathbb{C}}\Gamma(U,\mathcal{O}_U) \otimes_{\mathbb{C}} \mathcal{O}_V \to E\otimes_{\mathcal{O}_V} \dots \otimes_{\mathcal{O}_V} E.$$ There is a natural action of the finite symmetric group $\mathfrak{S}_n$ on the domain and the target, and $\phi^{\otimes n}$ is equivariant for these actions. In particular, there is a restriction homomorphism on the subrings of $\mathfrak{S}_n$-invariants, $$\phi^{\otimes n}_{\mathfrak{S}_n} : (\Gamma(U,\mathcal{O}_U)^{\otimes n})^{\mathfrak{S}_n}\otimes_{\mathbb{C}} \mathcal{O}_V \to (E^{\otimes n})^{\mathfrak{S}_n}.$$ Finally, the determinant above defines a homomorphism of $\mathcal{O}_V$-modules,
$$\widetilde{\text{det}}_F: (E^{\otimes n})_{\mathfrak{S}_n} \to \mathcal{O}_V.$$ Composing with $\phi^{\otimes n}_{\mathfrak{S}_n}$ defines a homomorphism of $\mathcal{O}_V$-modules, $$\text{det}(\phi):(\Gamma(U,\mathcal{O}_U))^{\otimes n})^{\mathfrak{S}_n}\otimes_{\mathbb{C}} \mathcal{O}_V \to \mathcal{O}_V.$$ By the algebra fact above, this is a homomorphism of commutative $\mathcal{O}_V$-algebras. This algebra homomorphism defines a morphism $V\to U^{(n)} \subset X^{(n)}$. These morphisms glue to a morphism $S\to X^{(n)}.$

**Edit. Proof of the Algebra Fact.** For certain choices of $R$, there are some funny commutative subalgebras $A$ of $E$, cf. my answer to the following question: Simultaneous triangularizability over a commutative ring. Thus, it makes sense to give a proof of the algebra fact. Actually it follows from a more general fact about $\widetilde{\text{det}}_F$ on the associative $F$-algebra $(E^{\otimes n})^{\mathfrak{S}_n}$.

Since $n!$ is invertible in $R$, there is a basis of $(E^{\otimes n})^{\mathfrak{S}_n}$ as follows. For every ordered $n$-tuple $(I,J) = ((i_1,j_1),\dots,(i_n,j_n))$ of pairs $(i,j)\in \{1,\dots,n\}\times \{1,\dots,n\}$, define $$\epsilon_{I,J} = \sum_{\sigma\in \mathfrak{S}_n} \mathbf{e}_{i_{\sigma(1)},j_{\sigma(1)}}\otimes \dots \otimes \mathbf{e}_{i_{\sigma(n)},j_{\sigma(n)}}.$$ The element only depends on $(I,J)$ up to permutation by an element in $\mathfrak{S}_n$. By direct computation, $\widetilde{det}_F$ is nonzero on this element if and only if $(I,J)$ is the graph of a bijection, i.e., $((\tau(1),1),\dots,(\tau(n),n))$ for some $\tau\in \mathfrak{S}_n$. In this case, denote the element by $\epsilon_\tau$. By direct computation, $\widetilde{\text{det}}_F(\epsilon_\tau)$ equals $\text{sgn}(\tau)$. Also $\epsilon_\tau \cdot \epsilon_\rho$ equals $\epsilon_{\tau\cdot \rho}$ for $\rho,\tau\in \mathfrak{S}_n$. Finally, if $\epsilon_\tau\cdot \epsilon_{I,J}$, resp. $\epsilon_{I,J}\cdot \epsilon_\tau$, equals $\epsilon_\rho$, then $\epsilon_{I,J}$ equals $\epsilon_{\tau^{-1}\cdot \rho}$, resp. $\epsilon_{\rho\cdot \tau^{-1}}$. Altogether, this implies that $\widetilde{\text{det}}_F(\epsilon_{I,J}\cdot \epsilon_{I',J'})$ is zero unless both $\epsilon_{I,J} = \epsilon_\tau$ and $\epsilon_{I',J'}=\epsilon_\rho$ for some $\rho,\tau\in \mathfrak{S}_n$. Moreover, in this case, $\widetilde{\text{det}}_F(\epsilon_\tau\cdot \epsilon_\rho)$ equals $\text{sgn}(\tau\cdot \rho)$. Since the sign of a permutation is a group homomorphism, it follows that in all cases, $$\widetilde{\text{det}}_F(\epsilon_{I,J}\cdot \epsilon_{I',J'}) = \widetilde{\text{det}}_F(\epsilon_{I,J})\cdot \widetilde{\text{det}}_F(\epsilon_{I',J'}).$$ Therefore, already the $R$-module homomorphism $$\widetilde{\text{det}}_F:(E^{\otimes n})^{\mathfrak{S}_n} \to R,$$ is a homomorphism of associative, unital $R$-algebras. Thus, the restriction to any commutative $R$-subalgebra of $(E^{\otimes n})^{\mathfrak{S}_n}$ is a homomorphism of commutative $R$-algebras.