It helps to view this in terms of the equation

Δ G° = -_nFE_°

Reversing the reaction reverses the sign on Δ G°, and therefore the sign on E°. The sign on the potential determines the direction in which electricity flows. By convention, a galvanic cell (spontaneous reaction) has a positive E°, while an electrolytic cell (non-spontaneous reaction) has a -E°. The magnitude of the potential stays the same, but a sign is added to signify the direction of current flow.

As far as adding half cells to determine the overall potential, this can be done because of the relationship in the above equation. You can see that E° depends on Δ G° and n. But for instance, if n is multiplied by 2, Δ G° will be multiplied by 2 as well (there will be twice as much Gibbs' energy because the reaction involves twice the electron transfer). This means that you can add half-cell potentials to get the overall cell potential. Be careful to note though that if you multiply a reaction by a coefficient that you don't multiply the potentials, as you would for Δ G°; the potential does not depend on the stoichiometric coefficients, only on what the cathode and anode are.

It helps to view this in terms of the equation

$$\Delta G^\circ = -nFE^\circ$$

Reversing the reaction reverses the sign on $\Delta G^\circ$, and therefore the sign on $E^\circ$. The sign on the potential determines the direction in which electricity flows. By convention, a galvanic cell (spontaneous reaction) has a positive $E^\circ$, while an electrolytic cell (non-spontaneous reaction) has a negative $E^\circ$. The magnitude of the potential stays the same, but a sign is added to signify the direction of current flow.

As far as adding half cells to determine the overall potential, this can be done because of the relationship in the above equation. You can see that $E^\circ$ depends on $\Delta G^\circ$ and $n$. But for instance, if $n$ is multiplied by 2, $\Delta G^\circ$ will be multiplied by 2 as well (there will be twice as much Gibbs' energy because the reaction involves twice the electron transfer). This means that you can add half-cell potentials to get the overall cell potential. Be careful to note though that if you multiply a reaction by a coefficient that you don't multiply the potentials, as you would for $\Delta G^\circ$; the potential does not depend on the stoichiometric coefficients, only on what the cathode and anode are.