$\DeclareMathOperator{\Pic}{Pic}$The following example is taken from [1, Example 5.8].
Let $S \subset \mathbb P^1 \times \mathbb P^2$ be a general smooth surface of bidegree $(2,3)$, so $S$ is a K3-surface. I want to understand how to compute the Picard lattice of $S$. First we know the Picard group of the surrounding space, $$\Pic(\mathbb P^1 \times \mathbb P^2) = \mathbb Z \cdot \mathcal O(1,0) \oplus \mathbb Z \cdot \mathcal O(0,1).$$ Choose effective divisors $X = \{pt\} \times \mathbb P^2$ and $Y = \mathbb P^1 \times \mathbb P^1$, which belong to the two linear systems of the two generators. So we may identify $[X] = \mathcal O(1,0)$ and $[Y] = \mathcal O(0,1)$. We can also calculate the intersections, namely $$[X]^2 = 0, \quad [X]\cdot[Y] = [\{pt\} \times \mathbb P^1], \quad [Y]^2 = [\mathbb P^1 \times \{pt\}],$$ and furthermore $$[X]\cdot[Y]^2 = 1, \quad [Y]^3 = 0.$$ We also have $[S] = 2[X] + 3[Y]$, and from this we obtain $$[X] \cdot [Y] \cdot [S] = 3, \quad [Y]^2 \cdot [S] = 2.$$ Since $\gcd(2,3) = 1$, this means that the restrictions $\mathcal O(1,0)|_S$ and $\mathcal O(0,1)|_S$ are $\mathbb Z$-linearly independent elements in $\Pic(S)$ and we have calculated that the rank $2$ lattice $\Lambda \subset \Pic(S)$ generated by them has intersection matrix $$\begin{pmatrix} 0 & 3 \\ 3 & 2 \end{pmatrix}.$$ However, van Geemen claims actually $\Lambda = \Pic(S)$, i.e. $\Pic(\mathbb P^1 \times \mathbb P^2) \to \Pic(S)$ is also surjective. Why is that the case?
[1] Bert van Geemen, Some remarks on Brauer groups of K3 surfaces.
One can apply the following version of Noether-Lefschetz theorem. It can be found in this paper.
One thing need to be highlighted here is that the theorem assumed the hyperplane section need to be very general, namely, complement of countable union of divisors in the moduli. Geemen only assumed K3 surface being general. I'm not sure if it is a typo, or I missed something.