Why is creating water exothermic?
You’re mistakenly conflating the bond dissociation energies between the diatomic components and the product molecule. If the bond dissociation energy of water is larger than the bond dissociation energy of the individual diatomic components, then the system is “lower energy” (preferred) in the $\ce{H2O}$ state and that “extra energy” in the atomic state will be released in the form of exothermic heating. Remember that elemental $\Delta H = 0$, and the water molecule has stronger bonds/lower enthalpy and would require more energy to dissociate than the elemental components (which is what your math shows), meaning water won’t spontaneously dissociate into the diatomic gases since that would be an endothermic process and require some external energy. As an aside, keep in mind that dissociation of bonds in a molecule like water won’t be consistent between atoms (after the first hydrogen bond has broken from an $\ce{H2O}$ molecule, the remaining hydroxyl group requires less energy to dissociate). In sum, if your product has a higher bond dissociation energy than your reactant, it’s at lower enthalpy and the net of the reaction will be a release of energy in the form of exothermic heating.