Transitivity is the fundamental property of all relations that we call "something something" order. Of course, an equivalence relation is also transitive, and in fact is also a preorder.
So, maybe, one can start from transitive relations, split them according to whether they are reflexive, irreflexive, or neither. (Obviously, there's nothing new in this taxonomy.) On the irreflexive branch one gets exactly the strict partial orders. On the reflexive branch one gets preorders and their specializations, namely, partial orders and equivalence relations.
On the third branch we find the riff-raff transitive relations, and I'm not sure anybody calls them orders. There are also preorders that are neither partial orders nor equivalence relations, of course. So, maybe one could adopt the definition that an ordering relation is a binary relation that is transitive and either reflexive and antisymmetric or irreflexive.
The only main difference from the definition you consider is that a relation that is transitive and antisymmetric, but neither reflexive not irreflexive, is not considered an order relation.
Totality (linearity) can be specified by saying that for all $a$ and $b$, if $a \neq b$, then either $a R b$ or $b R a$. This works for both reflexive and irreflexive relations. (Thanks to @mlc for reminding me to cover this detail.)
In this answer I do not provide simple examples of preorders, but enable you to find them on base of simple examples of partial orders.
You can just start with a partial order $\langle B,\leq\rangle$ and a surjective function $\nu:A\to B$.
Then $\preceq$ defined by: $$x\preceq y\iff \nu(x)\leq \nu(y)$$ is a preorder on $A$.
Actually every preorder can be described this way. If you start with some preorder $\langle A,\preceq\rangle$ then you can take $B:=A/\sim$ where $\sim$ is the equivalence relation on $A$ characterized by: $$x\sim y\iff x\preceq y\ \wedge y\preceq x$$
For $\nu:A\to B$ you take the natural function prescribed by $a\mapsto[a]$.
This preorder is not a partial order if and only if $\nu$ is not injective.
Best Answer
You have it backwards - every partial order is a preorder, but there are preorders that are not partial orders (any non-antisymmetric preorder).
For example, the relation $\{(a,a), (a, b),(b,a), (b,b)\}$ is a preorder on $\{a, b\}$, but is not a partial order.