Old Wernher got it right. Of course. It all makes perfect sense, at least the transit to the Moon part and the relative velocities.
I calculate 121 hours for half of an Earth orbit with a periapsis 400 km above the equator and an apoapsis 100 km (60 mi as he stated) above the far side of the Moon.
At apoapsis, the vehicle will be going about 0.2 km/s in the Earth frame. The Moon is going about 1.0 km/s around the Earth. So for the vehicle to encounter the Moon, it will need to be ahead of the Moon, and the Moon will catch up to it from behind at 0.8 km/s.
Therefore as the Moon approaches the vehicle (or as the vehicle approaches the Moon — same thing), the Moon will slow down the vehicle as it pulls on it from behind (and even reverse its motion), in the reference frame of the Earth. As the Moon passes, it will then be in front and pull the vehicle forward, speeding it up again.
Only in the frame of reference of the Moon will it appear that the vehicle speeds up as it approaches the Moon. You cannot think of this problem as the Moon and Earth being immobile with the vehicle flying around them. The Moon is moving, in fact much faster than the vehicle as it approaches.
The velocity at Earth closest approach is close to 11 km/s, so yes, as Wernher describes, the speed is slower at the Moon and faster at Earth. Much faster.
As already noted, the orbit's shape and speed is determined almost entirely by the Earth, with only a very small portion of the trajectory dominated by the Moon's gravity. So you can use a two-body (Earth and vehicle) Kepler orbit to approximate the trajectory to the Moon very well, ignoring the Moon's gravity.
If you like, you can then correct for the Moon as a perturbation using patched conics, considering the Lunar flyby as a hyperbolic pass with a $V_\infty$ of 0.8 km/s, the difference between the vehicle velocity and the Lunar velocity. In the Lunar reference frame, the initial and final velocities at $\infty$ have the same magnitude, but different directions. When that is converted to the Earth frame, it will result in a $\Delta V$ to the vehicle which will cause it to depart from the initial Kepler orbit about the Earth.
Here's what the trajectory actually ends up looking like in the Earth frame, where the trajectory is orange and the path of the Moon is blue (Earth is at 0, 0):
After that close lunar flyby on the far side, the vehicle is in a much larger orbit of Earth with an apoapsis almost ten times the distance of the Moon. So Wernher didn't have the post-flyby state correct. The orbit he showed is what would happen if the Moon weren't there when you got to apoapsis.