I do not know how to correctly interpret Hilbert's Irreducibility theorem with Galois group as my aim.

Here $K$ is a number field (or simply $\mathbf{Q}$).

Scenario 1: Take a field $L$ that is a finite Galois extension of $K(t)$ ($t$ an indeterminate) with Galois group $G$. Writing $L=K(t)[X]/(f(t,X))$ for an irreducible polynomial $f(t,X)\in K(t)[X]$, and taking a specialization $t=a\in K$ guaranteed by Hilbert we can see the Galois group descends and we get a $G$-Galois extension over the number field $K$ as $K[X]/(f(a,X))$.

I understand this situation well.

Scenario 2: Instead of a $G$-Galois extension we are merely provided with an irreducible polynomial whose SPLITTING FIELD has $G$ as Galois group, so the {\it degree of the polynomial can be less than the order of $G$.}

I took the following example from Malle and Matzat's book on Inverse Galois Theory. (Page 88, attributed to Beckman). (Instead of a general degree $n$ I take $n=3)$.

He claims $f(t,X) = X^3-3tX +2t \in \mathbf{Q}(t)[X]$ is irreducible with $S_3$ as Galois group. (of its splitting field).

For the special value $t=4$ we get the irreducible polynomial $X^3-12X+8$, but the discriminant is a square (of 72) and we get a cubic number field as splitting field and not the expected $S_3$ extension of $\mathbf Q$.

What mistake am I making in this scenario?

Instead of giving a degree $n$-polynomial in $K(t)[X]$ with $S_n$ as Galois group I would be more comfortable with an irreducible polynomial of degree $n!$ with $S_n$ as Galois group so that I can specialize that polynomial. Perhaps it is expecting too much.