I think the description in your first paragraph is perfectly reasonable, but I'll try to give a slightly more technical explanation.
If you hold a substance at temperature $T$, the phase that it adopts is determined by a trade-off between energy $E$, which it tries to minimize, and entropy $S$, which it tries to maximize. The actual phase at a given temperature is the one that minimizes the free energy $F = E - TS$.
Entropy is a measure of the disorder of the system, and is high in the gas phase, where the molecules are essentially moving freely and independently of each other, and lower in the liquid, where the density is high and the molecules have less freedom to move. But the energy is also higher in the gas phase, where the molecules have lots of kinetic energy and roughly zero potential energy, than the liquid phase, where they have less kinetic energy and negative potential energy (because of attractive interactions between the molecules, the “intermolecular bonds”).
At high $T$, the best way to minimize $F$ is for the substance to form a gas and hence maximize $S$. At lower $T$, it instead minimizes $F$ by forming a liquid, which reduces $E$. (At still lower temperature, it forms a solid, which has even lower energy and entropy.) At some intermediate temperature, which we call the boiling point $T_b$, the two effects balance out: the energy is lower by $\Delta E$ in the liquid phase and the entropy is higher by $\Delta S$ in the gas; if $\Delta E = T_b \Delta S$ then the two phases have the same free energy, and there is nothing to choose between the two. ($\Delta E$ is called the latent heat.) At this temperature, the two phases are in equilibrium.
So imagine starting with a liquid at $T < T_b$ and heating it up. As you put energy in, the molecules start to move a bit faster and to get a little further apart, increasing both their kinetic and potential energy. But the substance is still a liquid as long as you stay below $T_b$. Once $T$ reaches the boiling point, the extra energy goes into converting liquid into gas, which costs energy $\Delta E$, instead of increasing the temperature. Both liquid and gas are then at the same temperature (as they must be, to be in equilibrium), but the molecules in the gas phase have more energy. Finally, when all of the substance has become gas, the extra energy again goes into increasing the temperature.