What does an element-wise stabiliser mean? 
When a group acts on a set, a lot of important information about the
  action is contained in the stabilizers of particular elements. Recall
  that $G_{u_1,...,u_k}$ denotes the (elementwise) stabilizer in G of
  the elements $u_1, . . . , u_k$.

In this particular case we are talking about a group of automorphisms acting on a set of elements of a field. I believe the "action" of $G$ on the field is just applying an element of $G$ (an automorphism) to a element of the field. (Is this correct?)
But I have no idea what the first paragraph means in particular I have never come across the term elementwise stabiliser? Could it mean the set of all automorphisms of $G$ that map each element of $\{u_1,...,u_k\}$ to itself? Or maybe it is some kind of permutation on these $k$ elements?
 A: If $G$ acts on a set $X$ and $A\subseteq X$ and for some $g\in G$ the set $gA=\{\,ga\mid a\in A\,\}$ happens to equal $A$, the we say that $g$ stabilizes the set $A$. But it may still be the case that $ga\ne a$ for some (or even all) $a\in A$. Under the stronger condition that in fact $ga=a$ for all $a\in A$, we say that $g$ stabilizes $A$ element-wise.
Example: The reflection at a line stabilizes this line point-wise, but it also stabilizes any line perpendicular to it (but not point-wise).
A: so Let $G=Gal(L/K)$  so $G$ acting on $L$ is correct, and the
element-wise stabilizer of a set $ A=\{u_1,...,u_k\}$ is the
sub-group $\cap_{1\leq i\leq k}G_{u_i}$ that is all element of $G$
which fixe each $u_i$ , $i\in\{1,...,k\}$, it is a sub group of the stabilizer of $A$ ( $st(A)=\{g\in G\mid  g(u_i)\in A, \forall i, 1\leq i\leq k\}$).
A: Suppose $G$ acts on a set $S$ and $A$ is a subset of $S$. The set of elements in $G$ which fixes $A$ setwise is $$G_{\{A\}} := \{g \in G: A^g = A\}.$$  The set of elements in $G$ which fixes $A$ pointwise (aka element-wise) is  $$G_{(A)}:=\{g \in G: a^g = a, \mbox{ for all } a \in A\}.$$  Observe that the elements in this latter subgroup must fix each element of $A$  whereas the elements in the first subgroup are required only to permute the elements of $A$ among themselves. Fixing pointwise (ie fixing each point in the set) is a stronger condition than fixing setwise (ie mapping the set to itself). 
