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.
$\require{cancel}$For a problem with small coefficients like that, trial and error works. We have
$$f(x+y) = (x+y)^4 + x + y + 1 = f(x) + f(y) + 1$$ This implies
$$f(x+1) = f(x) + f(1) + 1 = 0 + 1 + 1 = f(x)$$
So if $x$ is a root of $f$, $x+1$ is another root of $f$. In particular $\beta+1$ is a root of $f$.
Besides $f(\beta^2) = (f(\beta))^2 = 0$, because the Frobenius is a field automorphism (as Jyrki Lahtonen told you in the comments). It follows that the roots are $\beta$, $\beta+1$, $\beta^2$, $\beta^2+1$.
Best Answer
If you want to continue the way you started, i.e. with $$x^3 + 2x+1 = (x-\beta) (x^2+ \beta x+ \beta^2 + 2)$$ you can try to find the roots of the second factor by using the usual method for quadratics, adjusted for characteristic 3. I'll start it so you know what I mean:
To solve $x^2+ \beta x+ \beta^2 + 2=0$, we can complete the square, noting that $2\beta + 2\beta = \beta$ and $4\beta^2=\beta^2$in any extension of $\mathbb{Z}_3$ (since $4\equiv 1$). $$x^2+ \beta x+ \beta^2 + 2=(x+2\beta)^2 + 2=0$$ But this is easy now since this is the same as $$(x+2\beta)^2 = 1$$ which should allow you to get the remaining roots.