Recall that F* is the group of units in F, and similarly for R*. The latter is a subgroup of the former. The quotient group G = F*/R* is the valuation group associated with R.
The map from F* onto G is a valuation of F. There may be several valuations. Each valuation ring R inside F implies a different subgroup R*, which produces a different homomorphism - a different valuation on F.
For a given ring R and valuation group G, take any two elements of G and pull them back to x and y in F. Either x/y or y/x is in R. Assume x/y lies in R.
Multiply x or y by any unit, anything in R*, and x/y still lies in R. It doesn't matter which x and y we select. It only depends on the elements of G. For any x and y in G, either x/y is in R, or y/x is in R, or both.
Use this relationship to build a partial ordering on G. Write y ≤ x if x/y is in R.
Clearly x ≤ x; the relation is reflexive. And if x ≤ y and y ≤ x then x/y and y/x are both in R, and are inverses of each other. They both belong to the kernel R*. Both x and y are in the same coset of R*, and represent the same element in G. If two elements are ≤ each other they are equal.
Finally show transitivity. Assume z ≤ y ≤ x, whence x/y and y/z are in R. This implies x/z is in R, and z ≤ x. We have a partial ordering.
Since R is a valuation ring, x/y or y/x is in R. Every pair of elements is comparable, and the partial ordering becomes a linear ordering.
Assume y1 ≤ x1 and y2 ≤ x2. This means x1/y1 and x2/y2 are in R. Multiply them together and (x1x2)/(y1y2) is in R. In other words, y1y2 ≤ x1x2. Multiplication in F, which defines the operation in the group G, respects order.
But what about strict inequality? Suppoes y1 < x1 and y2 ≤ x2. We know, from the above, that y1y2 ≤ x1x2. Suppose they are equal. Thus (x1x2)/(y1y2) is a unit in R. Also, x1/y1 and x2/y2 are in R, and the former is a nonunit. So a nonunit times an element of R gives a unit; this is impossible. Therefore y1y2 < x1x2, and strict inequality is preserved.
Remember that strict inequality is preserved, hence 0 < x implies x < 2x. (Add x to both sides.) The powers of x advance in valuation, and do not cycle back around to zero. Every nonzero element of G has infinite order, and G is torsion free.
If G is trivial then R is a trivial valuation ring. Such a ring is a field, nothing but units. If the ring is nontrivial, let x be a nonzero nonunit and note that the powers of x have increasing valuations, as described above. The powers of x also produce an infinite descending chain of ideals. Therefore a valuation ring with a nontrivial maximal ideal cannot be artinian.
Given a fraction x in F, write it in lowest terms and look at the exponent e on p. Remember that this exponent can be negative, if p is in the denominator. Now x is in R iff e is nonnegative, and x is a unit in R iff e = 0.
When two fractions are multiplied together, the corresponding exponents on p are added. (You need the fundamental theorem of arithmetic to make this rigorous, but it's pretty clear.) When looking at the exponents of p, multiplication in F becomes addition in the integers. The kernel is precisely those practions with e = 0, hence the quotient group, i.e. the valuation of F, is equal to Z, the group of integers under addition.
If x and y are fractions then y ≤ x iff x/y is in R, iff there are more powers of p in x than in y. As the fraction gets smaller, with more and more powers of p in the denominator, its valuation decreases. The fractions with negative valuation are the fractions that are not in R.