Let's take it from first principles. I am starting away from the rope of the question, but I hope that I will be able to paint a good picture of what happens. This will require getting back to basics unless we want to have generalities that don't really explain anything.
The tldr is that internal forces in a system and between a system and its environment (air for example) can "absorb" energy, that is some of the work that you perform on a system using a force that you control does not get transferred to kinetic energy or gravitational potential energy. But, this is the kind of "generalities" that I want to avoid in the first place.
Conservation of momentum in systems.
If you have a system of interacting elements and two elements 1 and 2 interact through a force, we know that the force exerted by 1 on 2 is equal in magnitude and opposite in direction to the force exerted on 2 by 1 (Newton's third law). Therefore, the momentum imparted on 1 by 2 is equal and opposite to the momentum imparted by 2 on 1 as F = dp/dt. This has for consequence that in a system, internal forces cannot change the momentum of the whole system (equal to the momentum of all the elements of the system), only external forces can. This also has for consequence that if you know the external forces exerted on a system, you know how the system will change in momentum, globally, as a function of time, whatever happens inside the system as the external forces are redistributed. That is, the rate of change in the sum of the momenta of each portion of the system is equal to the total force external exerted on the system, whatever happens inside the system.
Of note, momentum is a vector and each component of the vector is conserved. If you have two identical electrons coming towards each other, they will typically not move in the same direction they were moving after the interaction as they would have to be perfectly aligned to do so. They can interact even if slightly offset from each other. However, if the electrons had the same speed before interaction, they will have the same speed after the interaction, whatever the direction they are then going. The center of mass will keep on moving at the same speed it was going before the interaction.
Conservation of energy in systems.
At the microscopic level, energy is also conserved. However, energy is tricky. Mechanics is usually concerned either with a few particles interacting together, or with substances that are approximated as continuous. However, in the real world, stuff is made out of atoms. A very large number of atoms. When you get many particles (of the order of Avogadro's number ) interacting together, a lot of things happen. Local applications of energy can be redistributed among many atoms and in the end what started as a local application of force in a specific direction results in atoms moving in all directions. To get back to my two electrons, energy is a scalar. Even if the electrons change direction, their mechanical energy does not change.
This is where we need to be very careful and where energy "losses" happen. In mechanics, all this redistribution of energy among many particles and all these changes in direction can be so drastic that the object as a whole does not move in a single direction at the speed it "should" be going anymore. Instead, all of its atoms vibrate, but in different directions, and even change direction of motion (it all depends on the type of system). This is called temperature/heat and mechanics by itself doesn't know how to deal with this. Derivation of what happens exactly in such energy redistribution is extremely complex. In mechanics, we deal with this by saying that energy was "converted" to heat, when all of the energy stays in the object/system we consider, or "lost to friction" or other things, when our system interacts with the outside world and leaks a bit of its energy, as we like to use conservation of energy.
It can happen also that an object deforms under the application of a force. In these cases, atoms that were in equilibrium at a certain position relative to their neighbors will be moved to a new equilibrium position. Moving the atom required giving is energy, to break attraction, and then accepting energy from the atom, so that it comes to its new equilibrium. This deformation is a very good way of transferring energy in directions that differ from the original force application.
As final remarks, for completeness, in solids, atoms move about an equilibrium position and transfer energy with their neighbors all the time, very rapidly. Also, this is a classical explanation. In real life, things are even more messy. For example, objects radiate energy in the form of photons all the time.
Interaction between atoms in a system can redistribute energy in all directions. Even a force applied normally at the surface of an object can result in atoms in the object moving in all kinds of directions. Note that some ways of moving may preferred, and if energy dissipation is not too quick, the atoms will want to move in some ways that is coordinated, so that, for example, a metal object will ring. Once energy has been redistributed in all directions, we don't call it "kinetic energy" of the object as such, which is usually reserved for the movement of the center of mass of the object, or of its constituents "macroscopic" objects. You did not get rid of energy, you just redistribute it.
Note that however, there is no way to get rid of momentum, due to Newton's third law, unless you interact with your environment. That is, you cannot get rid of "macroscopic" momentum without outside interaction, while you can get rid of "macroscopic" energy without outside interaction.
A simple example.
Imagine you have two balls close to each other and take a third ball in your hands to hit both balls simultaneously, moving in a direction perpendicular to the line joining the first two balls "though the middle". Since you hit the two initial balls a bit from the side, they will get momentum both in the direction in which you are moving, and in the direction perpendicular to this direction. The center of mass of the two initial balls will start moving in the direction in which you exerted a force, but if you forget that the balls are also moving away from each other in the other direction, you will "lose" energy.
Even better, now join the two balls with a spring. After you finish hitting the two balls, they will both move in the direction of the initial force, and vibrate sideways relative to this force (imagine a very fast interaction with the third ball, has if hitting a drum). If you look the system from very far, you will not see this vibration and it will seem like you dissipated energy, but that's only because you forget about the sideways movement. Now, imagine that you balls move in space and that they can sometimes hit a smaller ball. They will impart momentum to this smaller ball and slowly lose energy. This is analogous to friction.
In the rope example, I would not think of "collisions" though. A rope is not a chain. A rope is made of fibers that ware weaved together. The problem statement says that the rope is coiled. To start from a coiled configuration to a straight configuration, at the end, the fibers in the rope will slip on each other and exert friction on each other. Also, lifting the rope will create deformations inside the rope. In the rope case, energy is lost by friction between the fibers. Depending on the details, some energy may be dissipated in the air (friction between rope and air molecules). Depending on the exact structure of the rope, I will concede that maybe collisions occur, but I would not consider this as a main energy dissipation effect.
What is wrong with approach 2
Conservation of energy works very well in problems concerning rigid bodies, as there is a theorem stating that the work done on the rigid body is equal to the kinetic energy given by the mass of the object moving at the speed of its center of gravity ("translation" kinetic energy)plus a term including the moment of inertial of the object relative to the center of mass and its angular speed about the same point "rotation kinetic energy". However, this is a theorem that applies only to rigid bodies. The rope in our problem is not a rigid body.
To start from an horizontal coiled rope to a purely vertical straight rope, a lot of things had to happen that require deformations.This is where the "missing" energy goes. Either mechanical energy has been lost to friction, or it has been stored in stretching of the rope. This latter possibility seems to be neglected though as the length of the rope seems unaffected by the whole process, at least to first order.
How could one store mechanical energy in similar problem? Well, imagine a rigid rod lying on a table. If you pick it up at one end and move this end vertically, at the end, after leaving the table, the rod will oscillate. If you then let go of the rod when it is vertical, it will rotate on itself. Here, total mechanical energy is conserved, but not linear kinetic energy. You store energy in the rotation degree of freedom of the rod.
What is confusing about approach 1.
Approach 1 is appropriate as there is no way of "dissipating" momentum imparted by the outside force unless external friction becomes significant. The problem is stated so that is it not. However, the solution given in the question leaves a lot unsaid, which has caused confusion. I would prefer to split (even an idealization here) the rope into 3 sections:
Section 1: what is on the table.
Section 2: what is vertical in the air moving at speed v.
Section 3: the transition between the two.
The solution leaves a lot of things unsaid that may be confusing.
Section 1: The normal force and gravity are exerted on this section, so why don't we care about them? Because by definition the normal is what balances gravity. These two forces are equal and opposite, so they do no work and impart no momentum to section 1. Also, there is no tension at the free end of 1 as it is assumed that the rope is coiled static on the table.
Section 3: This is where all the messy stuff happens and its existence is omitted. One presumes that the rope is moved from its end at constant V, but at the same time presumes that a piece of rope goes from 0 to V in a time dt, implying an acceleration. The fact is that there is a transition zone, where a piece of rope is accelerated from 0 to V. How is this possible?
If you look at a piece of rope (I mean, literally, go, take a rope and pick it up), that is supported from above with a part that lays on a table, and try to lift the rope a bit, you will see that the rope is curved between the vertical and horizontal parts, with an horizontal distance between the two. This means that a large displacement in the vertical portion (away from where the rope lays flat) can lead to a small displacement close to the table, where a small piece of rope starts moving. It's just like a lever. Speed close to the point of rotation is very small, even if speed away is large. The point of rotation here always move as the rope is unwound. As you continue lifting the rope, this part of the rope will get closer and closer to the vertical and pick up speed. Approach 1 makes the implicit assumption that section 3 is small and negligible, or that at least, section 3 has the same velocity profile at all times while the rope is lifted.
I would prefer a solution in which we state what the momenta of sections 1 and 2 is as a function of time (always zero for section 1), and say that whatever happens in section 3 is not important as it is small and stays more or less the same as we pull the rope. Then, adding the momenta of all 3 sections at time t and using F = dp/dt, we can get the force we exert and integrating over x will give work. The formula we obtain at the end will be the same, but this would end a lot of confusion as to why this works and avoid details about accelerating masses while having this mass having a constant speed and other confusing statements.
[EDIT] Storage of energy other than gravitational potential energy and translation kinetic energy in the rope
One thing I forgot to mention is that you can maybe temporarily store energy in the rope. Imagine you have a mass linked to a spring. You pull the end of the spring at constant velocity horizontally. The spring will elongate while the mass accelerates. However, once the mass reaches speed V, the spring still exerts a force and the mass overshoots. There will oscillations and energy is stored in the spring/mass system. You will need to exert a force that is larger than the force required to move the mass from 0 to its final speed in absence of a spring, so that you can do work on the spring. However, if there is friction, as there always is in real world situations, you will lose this energy eventually. My answer presumes that energy is lost quick enough that we don't accumulate much energy in the rope, so that it does not have any movement other than vertical at speed V just as the last piece of rope leaves the surface. In real life, this may not be the case. So, even conservative internal forces can in fact do work against you, not just friction. You can get more details of how this works in this simple system by looking for the treatment of harmonic oscillators. The right keywords are "forced (or driven) damped harmonic oscillator", and more particularly "step input" to such movements.