This is question $9.7$ from Stewart's Introduction to Galois Theory (third ed.).
I said that $\Sigma $ being the splitting field of $f$ over $K$ implies that $\Sigma = K (a_1, a_2,..., a_n)$ and that $K \subseteq L \subseteq K (a_1, a_2,..., a_n)$ implies that $L = K (a_1, a_2,..., a_r)$ for some $r \le n$. Hence, as $f$ splits over $\Sigma$ and $\Sigma = L (a_{r+1}, a_{r+2},..., a_n)$, $\Sigma$ is the splitting field of $f$ over $L$.
Is this correct/complete ?
No, this is not correct - intermediate fields cannot be classified so easily. For example, the field $ \mathbf Q(\zeta_7) $ is the splitting field of $ X^7 - 1 $ over $ \mathbf Q $, but it has the subfield $ \mathbf Q(\sqrt{-7}) $ which is not generated over $ \mathbf Q $ by any combination of roots of $ X^7 - 1 $.
The correct approach is the following: $ \Sigma $ is the smallest field extension of $ K $ over which some polynomial $ P \in K[X] $ splits. Since $ K \subset L \subset \Sigma $ and $ P \in L[X] $, $ \Sigma $ is also the smallest extension of $ L $ over which $ P $ splits, so by definition it is the splitting field of $ P $ over $ L $.