Forgive me if I misunderstood you, but it sounds like you expect the relay to transfer something to its "output" when it activates. This is not going to happen (unless you connect the switch contacts differently), since the switch contacts are completely isolated from the coil and everything else. This is clearly depicted in the circuit symbol for a relay:
simulate this circuit – Schematic created using CircuitLab
Note: the dotted line indicates a magnetic coupling, to physically "pull" the switch mechanism. It is not a current path, and doesn't imply that the relay switch contacts have any electrical connection to the coil in any way.
In the left circuit, when the relay "activates" the COM and NO contacts will be physically connected together, but since there's no external connections to them, they will effectively connect some nothing to more nothing.
On the right I hope it's clear that when the relay activates (the transistor switches "on", allowing coil current to pass), then the NO contact will become physically connected to COM, and consequently to the battery's positive terminal.
In your schematic it's not clear what else is connected to the relay contacts. One seems to be grounded (COM), but the other (NO) is Q0/6.4C, which is presumably where you are measuring voltage.
When the upper relay is off, its NO terminal potential is ambiguous, since I don't know what Q0/6.4C is, but when the relay activates, the NO terminal becomes grounded also. That terminal will have 0V.
With the relay activated, you can expect the NO and COM terminals to have the same potential (since they are joined), and if you measure the voltage difference between them, you will measure 0V.
With the relay un-activated, the measured potential difference between NO and COM will depend entirely upon whatever else is connected to node Q0/6.4C.