I had read that resistances only lose energy in the form of heat.
Correct.
Is that what the authors are talking about?
Yes, the author is just using confusing language.
Do they mean to say that this internal energy is first stored in the resistor in the form of internal energy and then dissipated as heat?
The author means that electrical energy is converted to heat in the material of the resistor.
This is a bit confusing because heat has some different properties as compared with the electrostatic charge energy stored in a capacitor or the magnetostatic flux energy stored in an inductor.
The electrostatic or magnetostatic energies are still in a form which can do work on other parts of the circuit.
Heat, on the other hand, is for the most part lost.
It's not entirely lost because you can build an engine based on heat differences; for example you can boil water to spin a turbine like in a nuclear power plant.
However, in most cases the heat energy is no longer a major part of the dynamics of the circuit itself.$^{[a]}$
This is the case because heat energy is the motion of the atomic ion cores in the material of the resistor vibrating around, whereas the circuit operates on the electrical energy of the electrons.
You can verify that resistors generate heat easily by observing an electric stove.
The stove top is just a big resistor, and it works by putting current through it to generate heat.
$[a]$: As mentioned by CuriousOne in the comments, heat in resistive elements actually does interact with the electrical dynamics of the circuit in the form of noise.
This was first discovered by J.B. Johnson and then theoretically explained by H. Nyquist.
The phenomenon is called Johnson-Nyquist noise.
Johnson-Nyquist noise is fascinating and critical in understanding the role of electrical circuits in thermodynamic systems, as explained in another SE post.