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This seems to indicate that the pin 6 part of the drive motor is driven to +12V properly, and that the Pin 5 part of the motor is not driven (considering the motor image above).
Considering the schematic the opposite side should drive Q13 through Q14. These are driven by a 7406 OC inverter (pin 11 = F H, pin 10 drives base Q14. This should work as follows:
F H is low (as we’re reversing). So Pin 10 should be open, and driven to a + voltage, causing Q14 to start conducting, causing Q13 to do the same.
Next image:
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Yellow: base of Q13, stays at 0 → this means the - side of the reverse action is not driven as we saw.
Blue: Pin 11 of E8 the 7406 (F H, input). This is a noisy LOW,
Purple: Pin 10 of E8, driving the transistors.
For some reason we do not get pin 10 of E8 high. Looking at the schematic there is a single circuit that controls the voltage on all outputs of E8 which is related to the speed control. The output of E15 pin 7 (called “Servo Amp”) is connected to a 2.4/2.4K network to all outputs. If this remains low it might cause this problem, so let’s take a look at its in- and outputs.
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Yellow: E15 pin 5 
, Blue pin 6 
, purple pin 7 (output).
It looks like the + line here is at quite a higher voltage than the - line - so why is the output so low?
Let’s try to cut R64 to remove the feedback loop and see if this makes the motor run.. This resistor is not numbered correctly (apparently) on the component overview; it can be found immediately under C46 between the two resistor arrays.
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Dark blue is now the basis of Q13. Cutting the resistor had no effect, sigh.
Checking E8 pin 10 again at least shows something (purple trace):
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Still a very low voltage,
Putting the yellow trace on the collector of E14. This should show the +12V voltage from the “other” side switching on, and it does:
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Next part is to find out why the output of pin 10 stays so low. What I tried:
Remove Q14. No effect.
Replace 7406 with a new one. No effect.
This requires a bit more thinking 8-/.