A preordered group is a group along with a preorder that makes addition monotone. As it is well known, a congruence in a group is equivalent to a normal subgroup. Everyone knows that a preorder on a group is uniquely determined by its positive cone, so lets define the predicate $Px \iff x\geq 0$. So I'd expect that a congruence $\cong$ on a preordered group would just add the requirement that if $x\cong x'$ then $Px\iff Px'$. But in most cases this seems like too strong a condition to ask. For example consider $\mathbb Z$ with the usual order. We can quotient it as a group by $2\mathbb Z$ and lift this to the category of preordered groups to obtain a morphism $\mathbb Z \to C_2$ where the order in $C_2$ is the indiscrete one. In fact we can do this for any $(G,N)$ where $G$ is a preordered group and $N$ a normal subgroup. But for the example I gave the relation induced by $2\mathbb N$ is not a congruence: we have $2\cong -2$ but $P(2)$ and not $P(-2)$. Therefore in most interesting cases these maps are not quotients since they are not given by congruences.
My approach is rather naive. I am aware there is a category theory definition of congruence which specializes well to varieties of universal algebra in the way I did it. Perhaps it doesn't work well when you add relations. The definition of congruences in category theory seems very technical to me so I'd like to ask if they agree with what I did here. And if not, is the example I gave actually a quotient?
If my examples are not quotients, is there a name for this construction?
The category-theoretic definition of congruence is really just "a reflexive, symmetric, transitive relation"; it looks complicated because it's phrased internally to a category. But in categories with reasonable products and pullbacks there's nothing much to it.
So, what are pullbacks in the category of preordered groups? If $A\to C\leftarrow B$ is a span in this category, then $A\times_C B$ is just the subgroup of $A\times B,$ with the product order, on pairs of elements sent to the same place in $C.$ (That is, pullbacks are created by the forgetful functor into preorders.) In particular, monomorphisms and products in this category are exactly what you think they are.
So, a relation on $A$ in the category of preordered groups is just a subgroup $R$ of $A\times A.$ It's a reflexive relation if it contains the diagonal subgroup $\{(a,a)\}.$ In this case we have $aRa'$ if and only if $eRa'a^{-1},$ so the relation is determined by its kernel, which is a normal subgroup. Symmetry is just symmetry ($R$ must be fixed by the automorphism of $A\times A$ that switches the coordinates.) Note this holds as soon as $R$ is reflexive, since $aa'^{-1}$ is in a subgroup if $a^{-1}a'$ is. And transitivity is just transitivity. Formally, it says that the projection from $R\times_A R\to A\times A$ sending $((a,a'),(a',a''))$ to $(a,a'')$ is contained in $R,$ but this means the usual thing. And again, it's implied by reflexivity, since a normal subgroup is closed under products. (All this happens for pretty much any kind of category of groups-with-extra-stuff, and more generally for any Mal'cev category. It's one of the key things that distinguishes stuff-like-group-theory from stuff-like-set-theory.)
Summarizing the above paragraph, a congruence in the category of preordered groups is nothing more than a normal subgroup $N\lhd G$ (with the induced preorder.) To take the quotient, equip $G/N$ with the smallest preorder which respects the multiplication and makes $G\to G/N$ an order-preserving map. In your example of $\mathbb Z/2\mathbb Z,$ this unsurprisingly produces the indiscrete preorder, since that's the only nontrivial preoder on $\mathbb Z/2\mathbb Z$ anyway.
There's a stronger notion of "congruence" on a preorder that would guarantee we can equip $G/N$ with the order in which $gN\le g'N$ if and only if $g\le g'.$ For this, we'd need to know that if $g_1g_2^{-1},g_1'g_2'^{-1}\in N$ and $g_1\le g_1',$ then $g_2\le g_2'.$ But on a preordered group, I believe this notion is almost vacuous. Setting $g_2=g_1=e,$ this would say that any element equivalent to a nonnegative element is nonnegative, which implies that $N$ is entirely nonnegative, and thus also entirely nonpositive. So it looks to me like this would only happen when quotienting by an indiscrete normal subgroup.