@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.