Consider as a toy example a discrete distribution with four outcomes $p_0,p_1,p_2,p_3$.
Start from $m=0$, which means $i=0$, and $|\psi_0\rangle=|0\rangle$.
The state for $m=1$ would instead be $$|\psi_1\rangle=\sqrt{p_0+p_1}|0\rangle+\sqrt{p_2+p_3}|1\rangle,$$
and finally for $m=2$ we'd get the target superposition $$|\psi_2\rangle=\sqrt{p_0}|0\rangle+\sqrt{p_1}|1\rangle+\sqrt{p_2}|2\rangle+\sqrt{p_3}|3\rangle.$$
To go from one step to the other, the evolution mentioned in the quote would be, at the $m=0$ step,
$$|0\rangle \to |0\rangle(\sqrt{p_0+p_1}|0\rangle+\sqrt{p_2+p_3}|1\rangle),$$
because $\alpha_0,\beta_0$ would be the sum of the probabilities of the left and right half of the full interval. In practice, the sum of the first and last half of the probabilities.
In the $m=1$ step, the evolution would instead look like
$$\sqrt{p_0+p_1}|0\rangle \to |0\rangle(\sqrt{p_0}|0\rangle+\sqrt{p_1}|1\rangle), \\
\sqrt{p_2+p_3}|1\rangle \to |1\rangle(\sqrt{p_2}|0\rangle+\sqrt{p_3}|1\rangle).$$
If we can implement both these evolutions it's clear we go from $|0\rangle$ to the target superposition.
As an addendum, let's consider how this evolution is performed.
Again, start at $m=0$ with $|\psi_0\rangle=|0\rangle$. The recipe, specialising the notation in the paper to our example, is to first implement the evolution
$$|0\rangle\to |0\rangle |\theta_0\rangle, \qquad \theta_0\equiv\arccos\left(\frac{p_0+p_1}{1}\right) \equiv \arccos(p_0+p_1),$$
and then rotate another qubit conditionally to $|\theta_0\rangle$, that is,
$$|\theta_0\rangle \to |\theta_0\rangle(\cos(\theta_0)|0\rangle+\sin(\theta_0)|1\rangle)
= |\theta_0\rangle (\sqrt{p_0+p_1} |0\rangle+\sqrt{p_2+p_3}|1\rangle).$$
In the next iteration, you'd start with for example $\sqrt{p_0+p_1}|0\rangle$. Work out $f(0)=p_0/(p_0+p_1)$ and $f(1)=p_3/(p_3+p_4)$, which are used to define $\theta_0\equiv \arccos\sqrt{f(0)}$ and $\theta_1\equiv\arccos\sqrt{f(1)}$, and separately implement the evolutions (I'm writing both of the above steps in one line here):
$$|0\rangle \to |0\rangle|\theta_0\rangle\to
|0\rangle|\theta_0\rangle\left(\sqrt{f(0)}|0\rangle+\sqrt{1-f(0)}|1\rangle\right)
= |0\rangle|\theta_0\rangle\left(\sqrt{\frac{p_0}{p_0+p_1}}|0\rangle+\sqrt{\frac{p_1}{p_0+p_1}}|1\rangle\right),
$$
or equivalently,
$$\sqrt{p_0+p_1} |0\rangle \to |0\rangle|\theta_0\rangle(\sqrt{p_0}|0\rangle+\sqrt{p_1}|1\rangle),$$
which is what we wanted. A completely analogous procedure gives
$$\sqrt{p_2+p_3} |1\rangle \to |1\rangle|\theta_1\rangle(\sqrt{p_3}|0\rangle+\sqrt{p_4}|1\rangle).$$
The main assumption required for this procedure to be doable, is that the functions $f(i)$ must be efficiently computable classically.
This procedure clearly looks very similar to the one discussed in Preparing a quantum state from a classical probability distribution, or at least some of the underlying ideas are. They might operate under different resource assumptions though.
