Picard group of surfaces can be worked "by hands" in the case of surfaces, for example ruled surfaces, which are $\mathbb P^1$-bundle over a curve $C$. When $C = \mathbb P^1$ this gives a family of surfaces called Hirzebruch surfaces, and are isomorphic to $\mathbb P_{\mathbb P^1}(O(n) \oplus O)$. More example of surfaces are worked in the notes online by Reid : available here.
You have an exact sequence $0 \to \mathbb Z \to Pic(X) \to Pic(X \backslash Z) \to 0$, where the first map is $1 \to [Z]$ which allows you to compute the Picard group of the complement of a curve in $\mathbb P^2$ for example.
In fact, you have an isomorphism $Pic(X) \cong A^1(X)$ for the case of smooth surface. Again for nice case (the precise word is "affine stratification") the Chow group is isomorphic to the cohomology group $H^2(X)$ (say if you work over $\mathbb C$). Which allows you to compute it in lot of case !
Finally let met explain to you my favorite example since it is very simple. If you don't know toric geometry, consider it as advertisement.
A toric surface is an algebraic surface which can be encoded combinatorially by a union of cone in $\mathbb R^2$. Example of toric surfaces are $\mathbb P^2$, $\mathbb P^1 \times \mathbb P^1$, cone over the Veronese curve, Hirzebruch surfaces ...
If you have such a surface you have an exact sequence $0 \to \mathbb Z^2 \to \oplus_{e \in E} \mathbb Z e \to Pic(X) \to 0$ where $E$ is the set of edges.
For example, when $X = \mathbb P^2$, there are 3 edges so the sequence is $0 \to \mathbb Z^2 \to \mathbb Z^3 \to Pic(\mathbb P^2)$ and you find again that $Pic(\mathbb P^2) \cong \mathbb Z$. You have four edges for $\mathbb P^1 \times \mathbb P^1$ so you find that $Pic(\mathbb P^1 \times \mathbb P^1) \cong \mathbb Z^2$, generated by $0 \times \mathbb P^1$ and $\mathbb P^1 \times 0$.
The best reference for toric geometry is the book of Fulton, but it is maybe a bit difficult. A good start is the note by Cox.
Notice that the toric picture is very nice : you also have a really handy (a bit longer to compute) description of the Cartier divisor, and in particular you can see very easily when $Cl(X)$ is different from $Pic(X)$ !
This cannot happen. Grothendieck proved that for any such surface, the natural map $\operatorname{Pic} \mathbb{P}^3\to \operatorname{Pic} \hat{X}$ is an isomorphism where $\hat{X}$ is the formal completion. So it suffices to prove that the cokernel of the natural map $\operatorname{Pic}\hat{X}\to \operatorname{Pic} X$ is torsion free. If $X _n$ denotes the $n^ {th}$ order thickening defined by the ideal $I^n$ where $I$ defines $X$, we have a natural exact sequence, $0\to I^n/I^{n+1}\to \mathcal{O}_{X_{n+1}}^*\to\mathcal{O}_{X_n}^*\to 0$. Taking cohomologies, we see that the desired cokernel is filtered by subspaces of $H^2(I^n/I^{n+1})$ and since we are over complex numbers, these are torsion free.
Best Answer
Huybrechts' book (1.4.1, page 18) gives an example in $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1$: