@Tobias – your counterexample is a good deal more natural than mine – I think what's going on here is that given any sets \\(Q\\) and \\(P\\), we can "lift" any partial order on \\(P\\) to a partial order on \\(\mathrm{Hom}(Q, P)\\) using the pointwise construction. We don't need any kind of order on \\(Q\\) to do this.

Now the order you're putting on \\(\mathrm{Hom}(P, P)\\) is just a special case of this. It depends on the order on \\(P\\)-as-codomain but doesn't care about any order on \\(P\\)-as-domain. Consequently composition is well-behaved on the left, but not on the right.

Re "since you only care about the first letter anyway, we can define multiplication by just keeping the first letter" – unfortunately I don't think this works. If we throw away the tail of the juxtaposition we no longer have \\(B \leq A\\) but \\(C \otimes B \nleq C \otimes A\\), because the latter expression is just \\(C\\) on both sides. So we do have a monoidal preorder after all, and we don't get the "one-sided" counterexample we're after.

Now the order you're putting on \\(\mathrm{Hom}(P, P)\\) is just a special case of this. It depends on the order on \\(P\\)-as-codomain but doesn't care about any order on \\(P\\)-as-domain. Consequently composition is well-behaved on the left, but not on the right.

Re "since you only care about the first letter anyway, we can define multiplication by just keeping the first letter" – unfortunately I don't think this works. If we throw away the tail of the juxtaposition we no longer have \\(B \leq A\\) but \\(C \otimes B \nleq C \otimes A\\), because the latter expression is just \\(C\\) on both sides. So we do have a monoidal preorder after all, and we don't get the "one-sided" counterexample we're after.