Suppose two ideals in the valuation ring R do not exhibit containment. This means there exist x (in the first ideal) and y (in the second ideal) such that x does not generate y and y does not generate x. Yet either x/y or y/x lies in R. This is a contradiction. Therefore, given any two ideals H and J, either H contains J or J contains H. The ideals form a chain via containment. We already know R is local, so the maximal ideal M sits at the top of this chain.
Conversely, assume ideals are linearly ordered in an integral domain R. Is R a valuation ring?
Let F be the fraction field. Let x/y be a fraction in F. Either x*R contains y*R, or y*R contains x*R. Either x generates y or y generates x. Either x/y is in R or y/x is in R. This holds for every x/y in F, hence R is a valuation ring.
Mod out by a prime ideal P, so the result is an integral domain. Two ideals in the image that do not contain each other pull back to ideals in R that do not contain each other. This is a contradiction, hence the ideals of R/P are linearly ordered, and R/P is a valuation ring.
Also, S inverse of R is a valuation ring. Since R embeds in S inverse of R, we are merely producing a larger ring inside F. This ring extension certainly contains x or 1/x, because R does.