So if I unplug it and place a jumper I should see something
No. Have a look at the schematic. The control has to see ground on one wire and open circuit on the other one at the same time. You can’t just jumper the switch.
EDIT: Sorry, this is a newer design than I thought.
The schematic shows the Lockout switch in the "Unlocked (Pilot Activated)" position.
Pin 2 on the switch (wire 168 Green) should have battery power when the key is ON.
Pin 1 (wire M927 Green) should also be powered. This wire goes all the way to the Pilot Shutoff Solenoid and from there to frame ground. Solenoid resistance spec is 33.75 ohms +/- 5%.
Pin 4 (wire 700 Pink) goes to Connector J1, Pin 39 of the ECM. It forms a loop with......
Pin 5 (wire H766 Yellow) goes to the Lift/Tilt Kickout On/Off switch, where it forms a loop back to Connector J1, Pin 70 of the ECM as wire C216-C52 Black.
That whole loop has to be continuous, so check continuity between Pins 39 (700 Pink) and 70 (C216 Black) of connector J1 of the ECM with the machine powered off and the connector unplugged from the ECM.
Both those loops are gonna have to work to get the Pilot system to work.
I bet you haven't looked at the Kickout Switch yet..?