Concrete examples of valuation rings of rank two.

Qiaochu's answer is sound in principle, but in practice one needs to be more careful with the definition of the ring. The quotient field $K$ of $A$ consists of formal Laurent series of the form

$$f=\sum_{r=-r_0}^\infty x^r\sum_{s=-s_0(r)}^\infty a_{r,s}y^s.$$

Here $r_0$ is an integer and for each integer $r$, $s_0(r)$ is an integer (depending on $r$). So these are the power series where the powers of $x$ are bounded below and for each integer $r$ the coefficient of $x^r y^s$ is zero for all $s$ below a bound depending on $r$. This complicated-looking condition ensures that the product of two elements of $K$ is also an element of $K$ (note that one cannot multiply two general Laurent series).

Then $A$ will consist of all such series with the additional conditions that $r_0=0$ and $s_0(0)=0$. The valuation of an element $f$ is the least $(r,s)$ under lexicographic ordering with $a_{r,s}\ne0$. Here the ordering is $(r,s)<(r',s')$ if $r < r'$ or $r=r'$ and $s < s'$.

A more high-brow interpretation of the condition for memebership of $K$ is that the support of $f$, the setof $(r,s)$ for which $a_{r,s}\ne0$, should be well-ordered, that is each subset of the support has a least element. (With repsct to this lexicographic ordering of course.)

By considering a version of this construction in $n$ variables one can construct explicitly a ring with a valuation of rank $n$.