Effective divisor on smooth projective variety

Yes, your reasoning for (ii) is correct.

(i) If $h^0(X,\mathcal{O}(D_0)) = 0$, then $D_0$ cannot be effective by your cited result. Namely if $D_0$ were effective, then there exists $0 \neq s \in \Gamma(X, \mathcal{O}(D_0))$ such that $D_0 \sim (s)_0$ and in particular $\Gamma(X, \mathcal{O}(D_0)) \neq 0$.

(ii) If $D$ is effective, then it has non-trivial global sections. This is the contrapositive of (i).

Indeed, Hartshorne also notes immediately after that proposition the following statement: Let $D_0$ be an effective divisor, then $|D_0| \cong (\Gamma(X, \mathcal{O}(D_0)) - \{0 \})/k^{\times}$ where $|D_0|$ denotes the complete linear system of $D_0$. Non-trivial global sections up to $k^{\times}$ correspond to the realization of $D_0$.

In general on nice enough schemes, invertible sheaves $\mathcal{L}$ admit non-zero rational sections $s$ such that $\mathcal{L} \cong \mathcal{O}(\operatorname{div}{s})$, Vakil 14.2.E. The sheaves corresponding to effective divisors are precisely the ones which admit such non-zero global sections.