Yes, the first quadrant is +V and +I so the output will source a current flowing out to a +ve voltage. Similarly the third quadrant is -V and -I so the output will source a current flowing out to a -ve voltage.
The other two quadrants are the tricky ones.
Quadrant 2 is +V and -I, so the output has to sink current flowing into the output from a +ve voltage.
Similarly quadrant 4 is -V and +I, so the output has to sink a current flowing into the output from a -ve voltage.
What you need is a push pull transistor pair connected between 0V and +ve power rail and the same from 0V to the -ve power rail, with the emitters all joined together. It isn't as simple as just an op-amp output, probably need steering diodes to stop transistors turning on when they shouldn't.
Think of it as a supply of +ve voltage, if the output voltage is lower than what is desired then source current. BUT if the output voltage is HIGHER than the desired voltage then it has to sink it. Similar argument for -ve voltage.