$$\frac{MH + NH}{OH} = \sqrt{3}.$$
Take point $K$ on segment $BH$ such that $BK = CH$. Draw segments $OK$, $OB$, $OC$, etc.

Since $O$ is the circumcenter of $\triangle ABC$, we have $\angle BOC = 2\angle A = 120^\circ$. Also, since $H$ is the orthocenter of $\triangle ABC$, we have $\angle BHC = 180^\circ - \angle A = 120^\circ$. Therefore $\angle BOC = 120^\circ = \angle BHC$, so $B$, $O$, $H$, $C$ are concyclic. It follows that $\angle OBH = \angle OCH$.
Note that $OB = OC$ and $BK = CH$, together with $\angle OBK = \angle OBH = \angle OCH$, so $\triangle BOK$ and $\triangle COH$ are congruent. Thus $\angle BOK = \angle COH$ and

$$\angle KOH = \angle BOC = 120^\circ, \quad \angle OKH = \angle OHK = 30^\circ.$$
$\triangle OKH$ is an isosceles triangle with base $KH$ of the form $120^\circ - 30^\circ - 30^\circ$, so
$KH = \sqrt{3}OH$. Since $BM = CN$ and $BK = CH$, we know $KM = NH$. Therefore
$$\frac{MH + NH}{OH} = \frac{MH + KM}{OH} = \frac{KH}{OH} = \sqrt{3}.$$