Transitivity of the Galois group on roots

Step 1: Since $f$ is irreducible over $K$ and $r_1,r_2\in R\subset L$ are both roots, there is a $K$-isomorphism $\varphi:K(r_1)\to K(r_2)$ with $\varphi(r_1)=r_2$, $\varphi|_K=\mathrm{id}_K$ — the basic property of minimal polynomials: $K(r_1)\cong K[x]/(f)\cong K(r_2)$ via the isomorphism sending $r_1\mapsto x\mapsto r_2$.

Step 2: $L$ is a splitting field of $f$ over both $K(r_1)$ and $K(r_2)$ (since $L$ is a splitting field over $K$, and $K\subseteq K(r_i)\subseteq L$). By the isomorphism extension theorem for splitting fields, $\varphi$ extends to a field isomorphism $\sigma:L\to L$.

Step 3: Since $\sigma|_K=\varphi|_K=\mathrm{id}_K$, we have $\sigma\in\mathrm{Gal}(L/K)$, and $\sigma(r_1)=\varphi(r_1)=r_2$. So for any two roots $r_1,r_2$ there is always some $\sigma$ sending one to the other — precisely the definition of transitivity.