Temperature characterizes the average energy of atoms/molecules, as per Boltzmann distribution:
$$
w(\mathbf{p},\mathbf{q})\propto e^{-\frac{H(\mathbf{p},\mathbf{q})}{k_B T}},$$
where the potential energy may contain interactions, i.e., it is not simply a sum of potential energies of separate atom/molecules in external field. Further, one can then prove the equipartition theorem that the average energy per degree of freedom is $k_B T/2$.
However, there is nothing in these statements about the temperature being the kinetic energy. This is the case when we consider an ideal gas, in which (by its definition) the interactions between the atoms/molecules are neglected and all the energy is the kinetic one.
Thus, the reason why the boiling water is not increasing its temperature, is because the energy goes into overcoming the potential keeping the molecules close to each other in the liquid phase.
Update (Correcting some of the misleading statements made above.)
Let us consider a liquid-gas phase transition for a system of $N$ particles described by Hamiltonian
$$
H(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)=\sum_{i=1}^N\frac{\mathbf{p}_i^2}{2m} + \sum_{i=1}^N\sum_{j=1}^{i-1}U\left(|\mathbf{r}_i-\mathbf{r}_j|\right)=\sum_{i=1}^NK(\mathbf{p}_i) + V(\mathbf{r}_1;..., \mathbf{r}_N)
$$
The Boltzmann distribution is then
$$
w(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)=
Z^{-1}e^{-\frac{H(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)}{k_BT}}=
Z^{-1}
\left(\prod_{i=1}^Ne^{-\frac{\mathbf{p}_i^2}{2mk_BT}}\right)\times
e^{-\frac{V(\mathbf{r}_1;..., \mathbf{r}_N)}{k_BT}}
$$
The partition function is given by
$$
Z = \left(\prod_{i=1}^N\int d^3\mathbf{p}_ie^{-\frac{\mathbf{p}_i^2}{2mk_BT}}\right)\int_{\Omega^N}\prod_{i=1}^N d^3\mathbf{r}_i e^{-\frac{V(\mathbf{r}_1;..., \mathbf{r}_N)}{k_BT}}=
\left(4\pi mk_BT\right)^{3N}Z_1
$$
If we calculate the average kinetic energy of a single molecule, the configuration integral over space cancels out, and it reduces to simple Gaussian integration:
$$
\left \langle\frac{\mathbf{p}_i^2}{2m} \right\rangle=
\frac{\int d^3\mathbf{p}_i \frac{\mathbf{p}_i^2}{2m} e^{-\frac{\mathbf{p}_i^2}{2mk_BT}}}{\int d^3\mathbf{p}_ie^{-\frac{\mathbf{p}_i^2}{2mk_BT}}}=
3\left\langle\frac{p_{ix}^2}{2m}\right\rangle=
\frac{3\langle p_{ix}^2\rangle}{2m}=
\frac{3mk_BT}{2m}=\frac{3 k_BT}{2}
$$
In other words, the average kinetic energy of the molecules is proportional to the temperature.
The above remains true even at the temperature where the system undergoes phase transition, as the result above is independent in the configuration integral $Z_1$. It is this configuration integral that undergoes change in phase transition: in the gas phase the volume filled with gas is that of the containing reservoir, $\Omega_g$, whereas in the liquid phase it is only the volume filled with the liquid, $\Omega_l$ (note that this volume can be located in different parts of the container or even be composed of many disconnected volumes.) Thus, the internal energy is different in two phases, depending on the volume over which we integrate:
$$
\left\langle H(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)\right\rangle=
\int\prod_{i=1}^N d^3\mathbf{r}_i\int_{\Omega^N}\prod_{i=1}^N d^3\mathbf{r}_i H(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N) Z^{-1}e^{-\frac{H(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)}{k_BT}}=
\frac{3Nk_BT}{2} + Z_1^{-1}\int_{\Omega^N}\prod_{i=1}^N d^3\mathbf{r}_i V(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N) Z^{-1}e^{-\frac{V(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)}{k_BT}}.
$$
Thus, the internal energy experiences a discontinuity across the phase transition (see also Why is the internal energy discontinuous in a first-order phase transition?, note that we are dealing here with a first order phase transition.) This discontinuity equals to the latent heat absorbed or produced during the phase transition.
Of course, in practice energy does not change discontinuously, but gradually, as heat is transferred or taken away from the substance. However, in this process the substance is not in thermal equilibrium, and needs to be treated using the methods of non-equilibrium statistical physics, see, e.g., Phase Transition in the Boltzmann–Vlasov Equation.
Related question (brought up in the comments by @GiorgioP): Why, exactly, does temperature remain constant during a change in state of matter?
