If $\Gamma$ is a totally ordered abelian group, and $\Delta\subset\Gamma$ is a convex subgroup (meaning if $\delta,\delta'\in\Delta$ and $\delta\leq\gamma\leq\delta'$ then $\gamma\in\Delta$), then we are supposed to be able to define a natural order on the quotient $\Gamma/\Delta$ (a source, for instance, here) via the rule
$$[\gamma]\le[\gamma']\iff\gamma\le\gamma',$$
where $[\gamma]$ denotes the class of $\gamma$ in $\Gamma/\Delta$. However, this definition doesn't make sense to me. If we took any $\delta>1$ in $\Delta$, then $[\delta]\le[1]$ (because the two are equal) but $\delta\not\le1$.
Is my definition of the order on $\Gamma/\Delta$ wrong? I also can't even prove directly that this even defines an order, so maybe I am missing something?
The definition of the order on $\Gamma/\Delta$ should say that $[\gamma]<[\gamma']$ iff every element of $[\gamma]$ is $<$ every element of $\gamma'$. Equivalently, $[\gamma]\leq[\gamma']$ iff some element of $[\gamma]$ is $\leq$ some element of $[\gamma']$.
You are right that, when $[\gamma]\leq[\gamma']$, you might not have $\gamma\leq\gamma'$; the "some elements" in the definition might not be the $\gamma$ and $\gamma'$ that you're given.