As David Roberts said, I post the  solution for preorders:

for two preorders $(X,\leq),\ (Y,\leq)$,  let $(X,\leq)\times_l(Y,\leq)$ the lexicographic preorder (priority to the first factor  $(X,\leq)$). 

We observe that a  preorder morphism $(A,\leq)\to (X,\leq)\times_l (Y,\leq)$  is a set map  $(f, g): A\to X\times Y$ such that $f: (A,\leq)\to (X,\leq)$ is a preorder morphism, and $g: A\to Y$ a set map (we can view it as a morphism $g: (A,\leq)\to ch(Y)$ where $ch(Y)$ is the chaotic preorder on the set $Y$ (dont exists the functorial analogy for orders)) and If $a\leq a'$ and $f(a)=f(a')$ then we have $g(a)\leq g(a')$.

Reciprocally if we have the data maked as: a couple of morphisms like $f: (A,\leq)\to (X,\leq)$, $g: (A,\leq)\to ch(Y)$ such that *"If $a\leq a'$ and $f(a)=f(a')$ then we have $g(a)\leq g(a')$"*, then we get the (unique) morphisms  $(f, g): (A,\leq)\to (X,\leq)\times_l (Y,\leq)$. 


This is easy to traslate in categorical setting: Let $U: Preord\to Set$ the forgetfull functo from the preorders to sets category, this has a right adjoint funtor $Ch: Set\to Preord$ (the chaotic preorder), now the phrase *"$a\leq a'$ and $f(a)=f(a')$"* is described by the categorical  relation  $r_l: R_l:=R_A(\leq)\cap P\subset A\times A$ where $R_A(\leq)$ is the  preorder relation on $A$ and $P$ the pullback of $f$ by the diagonal morphism $\Delta_X\subset X\times X$. The $X\times_lY$ is universal (initial) for the objects $A$ with two morphisms $f: A\to X$, $g: A\to Ch(U(Y,\leq))$ such that $(g\times g)\circ r_l: R_l\to Y\times Y $ as a (unique) factorization to the preorder relation  $R_Y(\leq)$ of  $(Y,\leq)$.


now I realized it was just an easy exercise in translation of categorical logic, that is from the the formula to  categorical diagrams.