In the matter of gravity vs. gravitational waves, I've always found it easier to think of (static) electric fields vs. EM waves.
Think of a static field "going out" from a charge. It isn't really going out. The field lines don't have "ends" that travel out at the speed of light. That's because they don't have ends at all. They all end at charges, and they stretch as far as needed. Charge is never created, it's always + - dipoles that are created (like electron-positron pair creation) so there's no problems with field lines that go out from an electron and go to the limits of the universe and end on some + charges out there. They've been stretched for that long, since the Big Bang. We don't think of how fast they move. They've been there since the beginning when the universe was small, and now they span it completely. A billion light years is nothing.
Similarly with static gravity. Mass-energy cannot just appear or disappear, so the field lines never have ends that have to move outward. They're always connected to mass and energy far away, and have had since the Big Bang to do so. There's no point in asking how fast static gravity moves. It's just "there" from here to the edge of the universe.
If you start to suddenly move, with respect to an already established static gravity or electric charge field-line, the field DIRECTION moves immediately with you, and so does the direction to the source. That's just Lorentzian relativity. The speed of light is not being violated. A source a billion light years away would suddenly start to look like it is moving, but that's because its field is already out to where you are, and the field where-you-are, tells you. It changes direction when you move. It responds immediately to relative uniform motion, via the mechanism of the field that is already extended to each.
But if the static or gravitational CHARGE moves (accelerates) then there is a "kink" or update that moves out from it at c. You don't see this at all from far away, until time d/c has passed. That's the EM wave or gravity wave. It's not a relative thing between source and viewer, because acceleration is not "relative" in relativity. You can't pretend that the observer accelerates and the source does not.
The Lienard-Weichert potentials for EM have two terms for this reason. One is the static one that depends only on relative velocity and points at the source (so long as relative velocity has been constant for long enough). The other one shows aberration (does not point at source), retardation, and is a disturbance in the field due to source acceleration (not observer acceleration).
The slow dance of Sun and Moon are a mix of both effects, in the near-field. The static effects point right at the sources, and are due to fields that already extend to infinity. They have no "speed." However, the second order effects due to small amounts of source acceleration (from orbital acceleration) are tiny, but they are genuine gravitational waves, and they move outward at speed c. They are retarded and would show aberration.
Does this mean the curvature contains energy? Does this mean "gravitational energy" can itself cause (more) curvature?
In short, that is a way of saying it. However, it should be taken with a grain of salt: no one is saying that you can find a stress-energy tensor for the gravitational field itself. Think of it as an intuitive view.
Pick a Schwarzschild black hole, for example. It is a vacuum solution of the Einstein Field Equations, so there is no matter anywhere in spacetime to bend spacetime. Yet, it is bent. Furthermore, if we compute the mass of the spacetime (for example, the ADM mass), we'll find a non-vanishing value. A convenient way of understanding this is that it is as if there was gravitational energy sourcing this bending. Notice I'm not saying there is a gravitational stress-energy tensor nor giving a coordinate-invariant, local definition of gravitational energy. Rather, I'm saying this is a way of interpreting in an intuitive manner what is going on.
Similarly, one can do this things to interpret, for example, redshifts. In a black hole, it is common to understand the redshift of a photon as it "climbs" out of the potential well: the photon loses energy to the gravitational field and is redshifted as a result, and we state is as if the gravitational field was gaining energy even though it does not have a stress-energy tensor.
Another example is with an expanding Universe. In Cosmology, energy is not generally conserved. However, there is a way of interpreting the equations of General Relativity and keep energy constant at cosmological scales: add on gravitational energy. This is comfortable because it allows us to interpret things in terms of constant total energy, but uncomfortable because gravity does not have a stress-energy tensor, and so on. Sean Carroll mentions it in one of his blog posts.
As a summary, Thorne is nowhere stating that there is a local, coordinate-independent notion of gravitational energy. The point is that it is sometime convenient to understand the nonlinear effects of General Relativity as being due to some sort of gravitational energy. To be fair, one can derive the Einstein–Hilbert action in this way (see e.g., Zee's Einstein Gravity in a Nutshell, Chap. IX.5 if I recall correctly).
Best Answer
Part 1: no, the amplitude decays as 1/r. The power or energy of the wave decays like the square of that, like 1/r^2 (one over r squared). This is the same as in electromagnetic waves, the gravitational wave amplitude (in General Relativity) components obey the same equations as that for electromagnetic waves for the electric or magnetic field. The reason is that although forces go like 1 over r squared, that can be also used to see how the electric field (or Newtonian gravitational field) changes with distance r, but it is true for STATIC or slowly changing fields, not true for radiating fields, eg, propagating waves.
To figure out propagating waves you need to figure out how the time change of the fields (electric, magnetic or gravitational, they give the same results) change. You need to derive the wave equations for those. It then comes out that the fields go like 1/r. The fields are the amplitudes, A. The power or energy in all cases those cases goes like A squared. That holds true for radially propagating electromagnetic fields and gravitational fields. For gravitational fields it is the metric of the spacetime in Einstein's General Relativity equations that define the fields. For gravitational waves, in the linearized case, the waves are represented as small changes, with amplitude A, of the background metric. In both cases it is the square of A that is the power or energy (and actually it is power or energy density, per unit area), which then goes like 1/r squared.
It is easy why it must be so. Realize that those energies are conserved as they propagate. So the total energy at a distance r, summed around the whole spherical wavefront (yes, consider them radiating away radially symmetrically) is then the power or energy density, multiplied by the area of a sphere. Since area = 4 x pi x r^2, the energy or power density must go like 1/r^2, so the total power or energy is the same at any spherical distance r. It's energy conservation that leads to that behavior.
Please be aware that these statements for the gravitational field are true for linearized gravity, i.e. For Einstein's equations, but where the waves or disturbances in the gravitational field, the metric of spacetime, is small. For very strong gravity and gravity changes things get much more complicated because Einstein's equations are nonlinear, gravity interacts with itself, and gravity is also be a source of a gravitational field. Things get complicated and you need some understanding of General Relativity to understand what that really means. If you are interested, there's plenty good online lectures and summaries (and bad ones also, careful).
Part 2: the decay of gravitational wave amplitude or power or energy density. Does it depend on time?
It might (remember that in General Relativity you can change coordinates and mix space and time)@ but in the linearized Einstein equations it does not, if you stick with the flat coordinate system as the reference frame. It'll depend on distance. And yes, then there is so called lensing, which is waves passing close to bodies some of the gravity and distort spacetime (in your terms, all of those together really just create the curvature of spacetime), and thus space near them, and thus a bending of the waves like lenses do. So waves going near them will bend, and typically travel a longer distance (but it all depends on the overall distribution of mass around their overall path), and arrive delayed. We can measure the so called lensing delays for electromagnetic waves and actually used it to measure the mass densities around various galaxies and galaxy clusters. We can even measure the mass densities of dark matter around them. We have not yet observed gravitational wave lensing, but it is expected and will be used to measure more details from bodies or objects that are not visible through electromagnetism.
So, yes gravitational waves will vary their amplitude and power with the longer lensing distances, and be delayed. The effects will be small, and the amplitude changes will be much harder to detect accurately than the timing delays. The space based gravitational interferometer to be launched in the next few years will be able to see the delays, not sure about the distance variations through the amplitude changes, but time delays determines distance anyway so we'll measure those distance changes also.