There are counterexamples to order isomorphisms of ordered fields being field isomorphisms, see Is the multiplicative structure of a totally ordered field unique?. However, Wikipedia suggests that for real closed fields order isomorphism implies isomorphism: "If the continuum hypothesis holds, all real closed fields with cardinality the continuum and having the $η_1$ property are order isomorphic. This unique field $F$ can be defined by means of an ultrapower... This is the most commonly used hyperreal number field in non-standard analysis, and its uniqueness is equivalent to the continuum hypothesis."
How exactly does one recover operations from order if $F$ is real closed?
What you quoted from Wikipedia is correct, but your more general statement, which you claim Wikipedia "suggests", is not. Two real-closed fields can be order-isomorphic without being isomorphic as fields. For example, let one of the two fields be the field of real algebraic numbers. Then choose a transcendental number, say $e$, and let the second field consist of all the real numbers that are algebraic over $\mathbb Q(e)$. These two fields are isomorphic as ordered sets, because all countable dense linear orders are isomorphic (a theorem of Cantor), but they are not isomorphic as fields because they have different transcendence degrees.