How to estimate the $h^0(L^{\otimes n})$ M. Reid has a theorem as following:
Let $X$ be a non-singular projective complex surface, $L \subset \Omega^1_X$ be a line bundle. If $h^0(L^{\otimes n})\geq2$ for some $n\geq 1$, then there is a moephism $f:X\rightarrow C$ of $X$ onto a nonsingular curve $C$ of positive genus, and a ramification divisor $D$ (of the map $f$) such that $L$ and $f^*(\Omega^1_C)(D)$ coincide as sheaves. 
How to derive the following statements？
(a) there exists a constant $k$ such that for all $n$ $$h^0(L^{\otimes n} )< nk$$
(b) $L$ is not in the positive cone of $NS(X)$
 A: I'm not entirely sure what you mean by a ramification divisor when a surface maps to a curve....
On the other hand, by the Bogomolov-Sommese vanishing theorem if $L\subset\Omega_X^1$, then the Kodaira dimension of $L$ is at most $1$. This implies both statements. 

Remark:
If by the ramification divisor you mean the union of the singular fibers, then the statement of the result you are citing implies that $L$ is contained in the pull-back of a line bundle on the curve and this implies the bound on the Kodaira dimension, so you don't need to use the Bogomolov-Sommese theorem. Part (a) is essentially that statement that the Kodaira dimension of $L$ is at most $1$ and that implies that it cannot be ample. Of course, it could still be a limit of amples, so it is not in the positive cone, but it may be on its boundary. 
A: (a) The ramification divisor $D$ is contained in some fibers of $f$, so we can write $L=f^*M(-E)$, with $M$ a line bundle on $C$ and $E$ effective. Therefore $h^0(L^{{\scriptscriptstyle\otimes} n })$ is bounded by $h^0(f^*M^{{\scriptscriptstyle\otimes} n })=h^0(M^{{\scriptscriptstyle\otimes} n })$ (projection formula).
(b) What do you call the positive cone?  We have $(L)^2\leq 0$ and $(L\cdot F)=0$ for  $F$ a general fiber of $f$, so certainly $L$ is not ample. On the other hand if $D=0$ $L$ is nef. 
