I think I found the solution: if $P$ is a place as in the equival. statement of my problem as above, suppose that $l(P)=\deg(P)+1$ (equiv. such a differential does not exist). Then there exists $z$ trascendental over $K$ whose pole divisor is $P$. Multiply a non-zero holomorphic differential ($g\geq0$) by a sufficiently large power of $z$ we get an holom. differ. whose associated divisor does not contain $P$.

exactly, that is what I meant... I used clifford´s theorem (as in the exercises of Stichtenoth book) to solve the problem, but I would like a simple proof which avoids that. Essentially is the fact that if $l(A)=\deg(A)+1$ in genus greater than zero then $A$ is principal, which I can´t prove. Is it possible to prove it without extending the base field to the algebraic closure, as it seems that you suggest? It seems that the result is contained in the book of Deuring, "Lectures in the theory of alf. func.", Lemma Lectures 10, §20, but I don´t understand the proof...

if $A$ is a divisor of $F/K$ such that $0\leq\deg(A)\leq 2g-2$, then $$l(A)\leq 1+\deg(A)/2$$ where as usual $l(A)$ is the dimension over $K$ of the vector space $L(A)=\{x\in F|(x)+A\geq0\}$.