It needs to be made clear what "reversible" means in this context. A process is thermodynamically reversible if it can be perfectly reversed (so it retraces all past macroscopic states in opposite order) by an arbitrarily small change in system conditions, such as temperature or pressure somewhere.
For example, air inside a cylinder with a movable piston pushed back by Earth's atmosphere and pushed/pulled by some other external force (a hand, a motor) can undergo a compression process or expansion process depending on which direction the piston is moving. When this process can be perfectly reversed by a very small change in the external force acting on the piston, the process is thermodynamically reversible.
This can only happen when the piston is moving very slowly, otherwise gas inside would produce eddy currents, pressure and heat waves, so-called non-equilibrium states that can't be made to be exactly retraced by some macroscopic action. So thermodynamic reversibility requires that processes are very slow.
Process being very slow is not always enough to be thermodynamically reversible, mostly because of friction. Consider what happens when the piston experiences large static friction on its outer edge. No matter how slow the piston moves, the friction will still be there, dissipating work into heat, thus heating up the cylinder and piston and the air layers close to them. Then it would be impossible to reverse the compression/expansion process exactly just by small change of force, because in both directions of motion, work is lost to heating up the cylinder. So to approach thermodynamically reversible motion, roughly speaking it has to be both slow enough to prevent non-equilibrium states, and friction has to be minimized.
First, what you understand about signs and $W$ is correct. There are two conventions to write the work: $\Delta U=Q+W$ or $\Delta U=Q-W$. In the first convention, it is the work given to the system, in the second convention, it is the work taken from the system. Usually, it is better to stick to a single convention in a course but your teacher does not seem to do this. With the convention $\Delta U=Q+W$, you always have $W_{irr}>W_{rev}$. I always use this convention.
Then you should force yourself to be very clear about the terms you are using. First there are state variables (or state functions, they are synonyms). A state variable is a property of the system at a given time, when you're looking at it. Imagine a box on the table with a gas in it. You can ask: what are the values of the state variables of the gas NOW. They are pressure, temperature, volume... and then energy, entropy...
Then you have a process. A process is a physical action on a gas. This is for example pushing the piston a certain way, heating the gas a certain way... In a process you have an initial state and a final state : that is the state of the system before you do the action and after. In both, the system has certain values for the state variables. The difference of one variable between the initial and final states is called a variation. You can write for example:
$$\Delta U=U_{final}-U_{initial}$$
For a process, you have properties of the process. They are the quantities of energy given to the system in the form of work and heat, written $W$ and $Q$. They depend on the process. They cannot be state variables, because it is about what happens to a system during an action. You cannot look at a gas in a box and ask "what is the heat or work of this gas now". You can ask "I pushed the piston for 5 minutes, what is the work given to the gas during these five minutes".
You can ask however, is the work a variation of "something". Is there a mysterious state variable, call it $X$, such that during any process, we would have:
$$W=\Delta X$$
This answer is no. It was believed in the eighteenth century that it was the case for $Q$. People believed that giving heat to a system was increasing it's "heat", and people called it the "calorific fluid". People believed that during a process, you had:
$$Q=\Delta C$$
where $C$ would be a state variable meaning "the amount of heat contained in the system : its calorific content". This is false.
Then we realized that "smooth" processes were special in some sense and we called them reversible processes. For a reversible process, the amount of work given to a system in a infinitesimal step of the process is always:
$$\delta W=-PdV$$
where $\delta$ just means it is an infinitesimal quantity. $d$ means infinitesimal variation and is the equivalent of $\Delta$ in the infinitesimal case.
Let us call $\delta W_{rev}=-PdV$. The non infinitesimal equivalent is:
$$W_{rev}=\int_{path} -PdV$$
We know that $Q$ and $W$ are not variations of state functions. Could $W_{rev}$ be ? Is there a state variable $X$ such as for all processes:
$$W_{rev}=\Delta X$$.
If it was the case, the reversible work would only depend on the initial and final states of a process. Because we would have:
$$W_{rev}=X_{final}-X_{initial}$$
But $\int_{path} -PdV$ is the area under the path when the path is represented in the $P,V$ diagram. It is clear that the area depends on the path and not only the initial and final states. Thus, $W_{rev}$ cannot be the variation a state variable $X$, whatever this variable is.
As a conclusion, neither work, heat, reversible work or reversible heat are variations of state variables.
Then what does this mysterious sentence means:
(Furthermore) the delivery of work (and of heat) is identical for
every reversible process.
I haven't read Callen's book, I don't know the context, I'm only guessing. This is my own interpretation of the sentence without the context.
That's where vocabulary matters. A process is not a path. A path is the sequence of states during a process. It is the curve, for example in the $P,V$ diagram. The path does not tell you what you are doing exactly with the system. A path may show that the temperature of your gas progressively increases during your process a certain way. But it does not tell you what you are doing with the gas: are you heating it with a heat bath? Are you spinning a powerful fan inside it? These are completely different actions. In one case, you give it heat. In the other case, you give it work.
But some quantities do not depend on the process, only on the path. Clearly it is the case for:
$$W_{rev}=∫_{path}−PdV$$
It means that whatever process you are doing, as long as the path is the same and the process is reversible, the work will be the same. Same for heat:
$$Q_{rev}=∫_{path}TdS$$
I would rephrase:
The delivery of work (and of heat) is identical for every reversible process having the same path.
A good video introductory course (clear and easy) is available here:
https://www.khanacademy.org/science/physics/thermodynamics/laws-of-thermodynamics/v/quasistatic-and-reversible-processes
It is not so easy to find a text explaining this, since it is mostly considered as implicit from the start. A good elementary course is available here: https://jfoadi.me.uk/stat_therm.html. The text uses exactly the same denotations as me. (Denotations like $W_{rev}$, $Q_{rev}$ seem to be only used with students, I tried to bridge the gap between these ideas and a more classical text like the one in the link)
Best Answer
Understanding entropy change is much simpler than it seems from your description of Callen. Here are the basics:
Entropy is a physical (state) property of the material(s) comprising a system at thermodynamic equilibrium, and the entropy change between two thermodynamics equilibrium states of a system depends only on the two end states (and not on any specific process path between the two end states).
For a closed system, there are only two ways that the entropy of the system can change:
(a) by heat flow across the system boundary with its surroundings at the temperature present at the boundary $T_B$; this is equal to the integral from the initial state to the final state of $dQ/T_B$ along whatever path is taken between the two end states.
(b) by entropy generation within the system as a result of irreversibility; this is equal to the integral from the initial state to the final state of $d\sigma$ along whatever path is taken between the two end states, where $d\sigma$ is the differential amount of entropy generated along the path..
Contribution (a) is present both for reversible and irreversible paths. Contribution (b) is positive for irreversible paths and approaches zero for reversible paths. For any arbitrary path between the two end states, the two contributions add linearly: $$\Delta S=\int{\frac{dQ}{T_B}}+\sigma$$
Determining the amount of entropy generation along an irreversible path is very complicated so, to determine the entropy change for a system between any initial and final thermodynamic equilibrium states, we are forced to choose only from the set of possible paths that are reversible in applying our equation. The reversible path we choose does not have to bear any resemblance to the actual path for the process of interest. All reversible paths with give the same result, and will also provide the entropy change for any of the irreversible paths.
Armed only with these basics, one can determine the change in entropy for a closed system experiencing any process, provided that application of the 1st law of thermodynamics is sufficient to establish the final thermodynamic equilibrium state.