Let $k_s|K|k$ be a tower of field extensions and $K|k$ be finite and separable ($k_s$ is a separable closure of $k$). There exists $\alpha\in K$ such that $K=k[\alpha]$. Then the splitting field $L$ of the minimal polynomial of $\alpha$ contains $K$ and $L|k$ is a finite Galois extension.
Now I think that the following should be true:
If $\sigma_1,\sigma_2\in\mathrm{Gal}(k_s|k)$ and $\sigma_1|_K=\sigma_2|_K$ then $\sigma_1|_L=\sigma_2|_L$.
No - that's false in general: if $G ={\rm Gal} (L/k)$, and $H$ the subgroup that fixes $K$, it is not always true that $H=1$, which is equivalent to what you are saying (because, for $\sigma_1$ and $\sigma_2 \in G$, $\sigma_1|_K = \sigma_2|_K$ is equivalent to $\sigma_1 = \sigma_2 h$, for some $h\in H$).
A concrete example: $k=\mathbb Q$, $K = k ( \alpha)$, where $\alpha^3 =2$. Then there are three roots, $\alpha_1 = \alpha, \alpha_2, \alpha_3$ (of the polynomial $f(x) = x^3 -2$, and $L$ the splitting field of $f$ over $k$). Identify $G$ with the symmetric group on the symbols 1,2, 3: the subgroup $H$ (of order 2) generated by the 2-cycle $(23)$ is the subgroup of $G$ that fixes $K$.
As the Galois group ${\rm Gal} (k_s/k)$ is the inverse limit of the Galois groups of the finite (Galois) extensions, your guess is not correct!