Why do we choose the standard total order on the integers?

The standard order is (up to isomorphism) the only total order on $\mathbb Z$ that makes it an ordered group under addition. So I'd expect it to be useful in situations where addition plays a role; these are probably most (though certainly not all, as Ryan Budney pointed out in a comment to the question) of the situations that arise in practice

Maybe the following can be an justification:

A linear order of a semigroup can be continued by one and only one way to a linear order of its semigroup of fractions (L. Fuchs, Partially Ordered Algebraic Systems, Theorem III.X.4).

Note. Here the semigroup can be neither commutative nor cancellative.

I assume (your) monoids are cancellative.

Then the pre-order you define is a (partial) order if and only of the monoid $M$ is reduced (i.e. has no invertible elements besides the neutral one).

For getting an order on the Grothendieck group $G$ say the element in $M$ are "positive elements" and define on $G$ the relation $g\ge h$ if $g-h \in M$. This extends the preorder of $M$; is transitive and reflexive; and it is anti-symetric if and only if $M$ contains no non-trivial invertible elements.

It is a total order if $g$ or $-g$ in $M$ for each $g \in G$; such a monoid $M$ is sometimes called a valuation monoid (in analogy with valuation rings, which have this property with respect to their quotient field, for multiplication of course; thus also these monoids are more frequently noted multiplicatively).