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The movable part is $|D-F|$, where $F$ is the fixed part. It has no fixed components. On a smooth algebraic surface its associated divisor is nef: for each curve choose a member avoiding that curve and use nonnegative local intersection multiplicities.

On a smooth algebraic surface, distinct integral curves have nonnegative intersection numbers, because their intersection multiplicities are nonnegative. For any member $E\in |D|$ of the complete linear system of a divisor, $E\cdot C=D\cdot C<0$. Hence $C$ must be a component of $E$; otherwise the intersection would be nonnegative.

Subtracting one copy of $C$ from every effective member gives precisely the effective members linearly equivalent to $D-C$. Conversely adding $C$ gives a member of $|D|$. Thus $C$ is a fixed component andEquivalently, multiplication by its defining section induces an isomorphism $H^0(X,{\mathcal{O}}_X(D-C))\stackrel{\sim }{\to }{H}^0(X,{\mathcal{O}}_X(D))$. The original PDF has this linear-system equality; the TeX loses the bars and reduces it to an uninformative divisor identity.

Because $|D|$ has no fixed components, choose two members $G_1,G_2$ with no common component. This uses the infinitude of the algebraically closed field and avoidance of finitely many proper linear subspaces of the section space. On a smooth algebraic surface, their intersection number is the sum of local intersection multiplicities. Since $G_1\cdot G_2=D^2=0$, they are disjoint. A basepoint of $|D|$ would belong to both, so $|D|$ is basepoint-free.

The resulting morphism has nonconstant image since $\dim |D|\ge 1$. Its image cannot be a surface: a generically finite morphism defined by $D$ would make $D^2$ the positive product of its degree and the degree of its image. Thus the image is a curve. Apply Stein factorization to obtain $g:X\to C$ with connected fibers and $g_{\ast }{\mathcal{O}}_X={\mathcal{O}}_C$, where $C$ is a smooth projective curve. There is a positive-degree line bundle $L$ on $C$ with ${\mathcal{O}}_X(D)=g^{\ast }L$. The projection formula for sheaves and Riemann-Roch theorem give $h^0(X,mD)=h^0(C,L^m)=m\mathrm{deg}L+O(1)$. Therefore