In general, there need not exist a surjective morphism $L\rightarrow C_L(\xi)=L_{\xi}$.
Take a Heisenberg Lie algebra $L$ of dimension $2m+1\ge 5$, with basis $x_1,\ldots x_m,
y_1,\ldots ,y_m,z$ and brackets $[x_i,y_i]=z$. The center $Z(L)$ is generated by $z$.
Assume that there is a surjective morphism
$$\phi:L\rightarrow C_L(x_1)=\langle x_1,\ldots ,x_m,y_2,\ldots ,y_m,z \rangle.$$
Then $\ker(\phi)$ is a non-trivial ideal of $L$, which must
intersect $Z(L)$ nontrivially, because $L$ is nilpotent. By dimensional reasons, $Z(L)= \ker(\phi)$. Hence $L/Z(L)\simeq C_L(x_1)$. The Lie algebra on the left hand side is abelian,
the one of the right hand side not (here we need $m\ge 2$). This is a contradiction.

Concerning the structure of centralizers, this is a quite general question, with no clear
answer. However, the structure of $L$ can depend quite a lot on the centralisers.
As an example, if all centralisers of nonzero elements are abelian, then $L$ cannot be nilpotent. Furthermore there is the following interesting result of Jaikin-Zapirain:

Let $L$ be a nilpotent Lie algebra. Then the nilpotency class of $L$ is bounded in terms of
$$\Delta (L)=max \; cd(L)-min \; cd(L),$$
where $cd(L)$ denotes the set of dimensions of centralisers of nontrivial elements of $L$.