If you have access to the answer and this is incorrect, please let me know. I personally get
$$
v = \sqrt{\frac{2gx}{3}}
$$
which is the non temporal solution (v as a function of x) to
$$
g\ddot{x} = \frac{d}{dt}\left( xv \right)
$$
which is what I get as the Newton's 2nd Law equation for this system assuming the chain "enters with zero velocity." Please let me know if this is off base.
The resulting $x$, $v$ relation can actually be integrated too, giving the temporal answers:
$$
x = \frac{1}{6}gt^{2}
$$
and
$$
v = \frac{1}{3}gt
$$
Clearly the net effect of this looks like the local acceleration $g$ is reduced to a third the true value. Losses can be verified by calculating potential and kinetic energies with distance:
$$
\Delta PE = -gmh_{com} = -g \rho_{L}x * \frac{x}{2} = -\frac{g\rho_{L}x^{2}}{2}
$$
$$
KE = \frac{1}{2}mv^{2} = \frac{1}{2}\rho_{L}x * \frac{2gx}{3} = \frac{\rho_{L}gx^{2}}{3}
$$
where $\rho_{L}$ is the linear density and $h_{com}$ is the distance to the mass center of the moving portion of the chain.
The losses are then the balance of these changes, since they should sum to zero. For this I get
$$
E_{Loss} = \frac{\rho_{L}gx^{2}}{6}
$$
The source of the loss can be conceptually understood by viewing each differential segment of chain at the top entering the falling line of chain as a fully inelastic collision where one body was initially stationary. As you know, when two bodies collide and stick together, there is mechanical (kinetic) energy loss. The most energy loss occurs when a moving tiny body collides and sticks to a massive body stationary, since virtually all the kinetic energy is lost. This is like a pebble landing on the earth. Even in the reverse scenario, there are still losses, although less. Thus, the falling chain is the large body, and it "impacts" the "small body" differential segment that "sticks to it" (since they are now moving with the same velocity).
I will play with it a bit, but I bet that these equations of motion and impact analogy can actually predict the same losses.
Note: regarding the first example, there are no losses in that system because the entire chain is released at once and subject to gravitational force. The entire chain begins accelerating immediately and the force from the constraint at the top is smooth and only acts to redirect the motion. When it reaches the bottom, the "whiplash" is basically a big inelastic collision. So you can think of the second example as having "differential whiplashes" at every step of the way, but the those are in the opposite sense. They speed up stationary chain rather than stop moving chain. Nevertheless, they are inelastic collisions.
Further Note: In Johnston and Beer (that one is my personal favorite) they have a problem like this in the Systems of Particles chapter where they compare your second set up to one where the chain is laid out straight (horizontally) on top and moving as a unit. The assumption is probably supposed to be that the chain redirects with no losses at the hole. In this case, the chain enters the system with the same velocity as the system and so does not need to be sped up. According to my description here, there is no "collision" as it enters. In this case, it probably conserves energy, the book does not have a solution to that one but I bet that is the point to having the two set ups.