Let me do the case where $K={\bf Q}$. Let $L_f,L_g$ be the splitting fields. Then $L_f\cap L_g$ must be a subfield of both $L_f$ and $L_g$. So, what are the subfields of those splitting fields?
Well, $L_f$ has three real cubic subfields and one real quadratic; $L_g$ has one real cubic subfield, two imaginary cubic subfields, and an imaginary quadratic subfield. So if the intersection isn't $\bf Q$, it must be the real cubic subfield. But then we have a single cubic subfield that has three real embeddings and also two complex embeddings, and that's absurd. QED.
I don't think I made much use of the hypothesis $K={\bf Q}$; this argument ought to work more generally.
EDIT: More generally, if $L$ and $M$ are splitting fields of irreducible cubics over $K$ and are not equal then their intersection can't be of degree 3 over $K$; it can only be $K$ or the quadratic extension given by adjoining the square root of the discriminant (assuming the discriminant is not a square in $K$). That quadratic case can happen, as non-isomorphic cubic fields can be generated by polynommials with the same discriminant. Over the rationals, there are two non-isomorphic cubics with discriminant -1836, also three with discriminant 22356.