Let $E$ be a splitting field over $F$ of some $f(x)\in F[x]$, and let $K$ be a splitting field over $E$ of some $g(x)\in E[x]$.
If $\sigma \in Gal(K/F)$, then $\sigma |_E\in Gal(E/F)$.
Let $\lambda=\sigma|_E$.
Since $\sigma$ fixes $F$, we have $\lambda$ fixes $F$.
And also since $\sigma$ is an injection, so is $\lambda$.
So the things left are to prove that $\lambda(E)=E$.
Let $E=F(\alpha_1,\dots,\alpha_n)$ where $\alpha_i\in E$ are roots of $f(x)$.
Since $\lambda$ fixes $F$, $\lambda(\alpha_i)$ are also roots of $f(x)$.
Hence $\{\lambda(\alpha_i)\}\subseteq E$.
Let $x\in E$. Then $x=\sum_{i,j}k_{ij}\alpha_i^j$ where $k_{ij}\in F$.
Hence $\lambda(x)=\sum_{i,j}k_{ij}\lambda(\alpha_i^j)\in E$.
Hence $\lambda(E)\subseteq E$.
Is my last paragraph correct?
And I have no idea how to prove that it is surjection because $\lambda$ does not necessary permute $\alpha_i$.
I still haven't use the fact that $K$ is a splitting field of $E$.
Recall the definition of splitting field:
In the second paragraph, you have shown that $\lambda(E)$ contains all the roots of $f(x)$. Also since $F$ is fixed by $\lambda$, $F\subseteq \lambda(E)$. And as you have shown above, $\lambda(E)\subseteq E$. Hence from the definition it follows that $\lambda(E)=E$.
Your last paragraph contains a minor error. An arbitrary element $x\in E$ should be written as a linear combination of arbitrary powers of all elements $\alpha_1,\ldots,\alpha_n$.