Note: Your polynomials are still polynomials in one variable. It's the coefficients of the polynomial that are rational functions in several variables, but these coefficients are constants of your field.
To show that $F(u_1,\ldots,u_n)$ is Galois over $F(s_1,\ldots,s_n)$, it suffices to show that it is the splitting field of some separable polynomial with coefficients in $F(s_1,\ldots,s_n)$. Let $f(t)\in F(s_1,\ldots,s_n)[t]$ be the polynomial
$$f(t) = t^n - s_1t^{n-1} + s_2t^{n-2} + \cdots + (-1)^ns_n.$$
Since
$$f(t) = (t-u_1)(t-u_2)\cdots(t-u_n),$$
it follows that $F(u_1,\ldots,u_n)$ is indeed the splitting field of $f(t)$ over $F(s_1,\ldots,s_n)$.
Note also that $f(t)$ has no repeated roots, since the $u_i$ are pairwise distinct indeterminates, so $f(t)$ is separable. Thus, $F(u_1,\ldots,u_n)$ is the splitting field of a separable polynomial over $F(s_1,\ldots,s_n)$, and so it is a Galois extension of $F(s_1,\ldots, s_n)$.
To finish off, we need to show the Galois group is precisely $S_n$.
Note that $S_n$ acts on $F(u_1,\ldots,u_n)$ by fixing $F$ and letting $\sigma$ map $u_i$ to $u_{\sigma(i)}$; these automorphisms leave $F(s_1,\ldots,s_n)$ fixed, and they are pairwise distinct, so this proves that $G=\mathrm{Gal}(F(u_1,\ldots,u_n)/F(s_1,\ldots,s_n))$ contains a subgroup isomorphic to $S_n$. On the other hand, an element of $G$ is completely determined by what it does to $u_1,\ldots,u_n$, and it must map $u_i$ to a root of $f(t)$ above, hence elements of $G$ act as permutations of $u_1,\ldots,u_n$, showing that $G$ must be isomorphic to a subgroup of $S_n$. These two conclusions tell us that $G$ must in fact be isomorphic to $S_n$, as desired.
Another method is to note, as KCd remarks, that $S_n$ acts on $F(u_1,\ldots,u_n)$, hence, $F(u_1,\ldots,u_n)$ is Galois over the fixed field of $S_n$, with Galois group $S_n$. The fixed field of $S_n$ under this action is precisely the field of symmetric rational functions, which by the Fundamental Therorem of symmetric functions is generated over $F$ by $s_1,\ldots,s_n$; that is, the fixed field is $F(s_1,\ldots,s_n)$, which shows that $F(u_1,\ldots,u_n)/F(s_1,\ldots,s_n)$ is Galois with Galois group $S_n$, as desired.
Best Answer
You have the field $L$ with a faithful action of $S_n$. You have $F$, the fixed field of $S_n$ and $E$, the fixed field of $A_n$.
Recall the theorem of Galois theory that states that if $G$ is a finite group of automorphisms of a field $L$, and $L^G$ is its fixed field, then $L/L^G$ is a finite Galois extension with Galois group $G$. In particular, $|L:L^G|=|G|$.
in this example, $|L:F|=|L:L^{S_n}|=n!$ and $|L:E|=|L:L^{A_n}|=n!/2$. Therefore $|E:F|=2$ and so $E=F(\xi)$ where $\xi$ is any element of $E$ that's not in $F$. For instance, $\xi=\prod_{i<j}(x_i-x_j)$.