The way I understand the setup: The block is initially at rest at some height above the spring. The spring is initially at rest oriented vertically with one end on the ground and at it's natural length (though if its not a massless spring it would compress somewhat due to its own weight). Then, the block is dropped, it lands on the spring compressing it and at some time the maximum compression is reached.
At maximum compression the block is momentarily stationary (between moving down and back up again) and the spring should be stationary as well (though I expect there would be some vibrations). My point is initially and at maximum compression there is no kinetic energy. Overall the initial gravitational energy is converted to elastic potential energy.
With a massive spring it becomes a bit more involved. The spring will compress to some extent under its on weight and as it is compressed its centre of mass will move down reducing its gravitational potential energy.
To understand your situation, you need to rethink your understanding of the definition of the potential energy.
The best definition for the potential energy is: Potential energy is a measure of work done by a potential force upon the object moving while it is experiencing that force. The change in the potential energy of a body between two positions is equal to work of the potential force upon movement of the body between those two positions.
So, when you specify Potential energy of the body, it is always the energy of a specific [conservative!] force (or a combination of conservative forces), and with respect to some point of reference that you choose as a zero level of that potential energy. (Note that the position is also with respect to the other body that produces the force.)
When you have a single solid body, i.e. there are no external forces acting on it, you do not have any potential energy defined.
So, the answer to your question: in case of a single solid body the potential energy is undefined.
Saying that something that is undefined is zero is not strictly correct.
The case is rather different when the body is not solid, i.e. when its shape is not constant. E.g. if the shape of an elastic body can be changed (e.g. the body can be compressed), - then you have a potential energy related to that deformation.
In that case, you can consider the body as a combination of different parts ("points" of very small mass) of it that are acting on each other with forces. If the object is fully elastic, each of those forces is potential, and then you can associate potential energy with each of them. Accounting for all of them individually is rather difficult, so, an average potential energy for the entire ensemble of these portions (points) is introduced. In this, fully elastic case, this potential energy is a unique function of the shape of the body (i.e. of the relative positions of all the points within that body).
Some other typical examples where the confusion similar to the OP's occurs is the potential energy of a charged particle (e.g. electron) in an electric field. It looks like there is no other "body". In reality, there is a field (electric), which effectively is the representation of another body(ies) acting on this charged particle. Note that in case of gravity, one can define a gravitational field, thus "hiding" the object that creates it.
Best Answer
Where we define the potential energy to be $0$ in classical mechanics is arbitrary. All that matters is the change in potential energy. Since you are just learning this stuff I will assume you are in an algebra based physics class, so I will avoid using calculus here.
Potential energies are nice because they tell us how much work is done by a conservative force. More specifically, the work done by a conservative force is given by $$W_{cons}=-\Delta U$$ where $U$ is the potential energy associated with that conservative force. This is useful because we also know that the net work done on an object determines its change in kinetic energy $$W_{net}=\Delta K$$
So, if we consider your case where we just have one conservative force acting on the object, we can conclude that $$\Delta K=-\Delta U$$
And so we see here that the only thing that determines how the motion of our object changes is just the change in potential energy. If we define the zero-point to be at the top of the cliff, then as the object falls its kinetic energy will grow and its potential energy will decrease and be negative when it hits the bottom of the cliff. If we define the zero-point to be at the bottom of the cliff then as the object falls its kinetic energy will grow and its potential energy will decrease to $0$ when it hits the bottom of the cliff. In either case the same thing happens because we have the same change in potential energy.
Typically in introductory physics classes we just consider the ball being in a uniform gravitational field and we don't even consider the Earth. However if you want to include the Earth in your system then when the ball falls the Earth will actually move slightly upwards to meet the ball. It will still be the case though that the work done by gravity on each object is related to the change in the potential energy of each object, which results in a change of kinetic energy.