Also your calculation is not correct: You add another vertical 4,200 kph of free fall starting from apogee, but didn't factor in that of course the vehicle also slows down as much in the 2 minutes before reaching apogee. Since MECO is at about 65 km height at 8,300 kph, it will also have 8,300 kph at 65 km height when falling down again (minus a little atmospheric drag).
You also can't just add 8,300 kph horizontal (and it's not actually all horizontal, look how quickly the altitude is rising around MECO) and 4,200 kph vertical speed. Those are not scalar values but vectors, so you need to apply pythagoras for vector addition. 8,3 km/s horizontal plus 4,2 km/s vertical speed is just 9,3 km/s total speed.
