On a K3 surface, let $D$ be an effective divisor with $D^2\geq0$. Let $$D\sim D'+\Delta$$ be its decomposition in moving part and fixed part, respectively. Let $\Gamma$ be a prime component of $\Delta$ such that $D'\cdot\Gamma>0$. Assume $D'$ is irreducible. Then $$D'\cdot\Gamma=1$$
(This is remark 2.7.3 in Saint-Donat's article on K3 surfaces (p.611). He says that it is a straightforward consequence of 2.1 (i.e. Riemann-Roch) and 2.2 (which yields $h^1(D')=0$))
Honestly, I don't understand. I would be grateful if somebody could explain me why this holds.
By definition of the fixed part, we should have $h^0(D)=h^0(D')$. Let's show that implies the claim.
First assume that $\Delta$ consists of a single curve $\Gamma$ which intersects $D'$. (It must be a (-2)-curve.)
By the facts you mentioned, $h^0(D')=2+\frac12 (D')^2$, whereas $h^0(D) \geq 2+\frac 12 D^2$. But now
$$ D^2 = (D' + \Gamma)^2 = (D')^2 + \Gamma^2 + 2 D'\cdot \Gamma.$$
So if $D' \cdot \Gamma >1$ we get $D^2 > (D')^2$, hence $h^0(D) > h^0(D')$, which is a contradiction.
If now $\Delta$ consists of more components, just take one of them $\Gamma$ which intersects $D'$. Then $h^0(D) \geq h^0(D' +\Gamma)$, so the argument above still works.