The opamp's output is constrained by the power supply potentials, so you'll never get an output that falls outside that range. That's written \$-2V < V_{OUT} < +2V\$, in your example case.
This has nothing to do with what input potentials the device can tolerate, because the circuitry that deals with inputs is physically different from the circuitry that produces the output. They are separate "systems" within the IC, with a lot of other "systems" in between.
Acceptable input potentials are defined by the circuitry that handles them, and the range may or may not be different from the range of possible output potentials. There are good reasons to want the inputs to tolerate anything the output can produce, though; take a look at the voltage follower:
simulate this circuit – Schematic created using CircuitLab
This is a very common configuration, used to "buffer" some input signal, to provide a strong copy of it that's able to drive loads which require more current than the original source has to offer.
The output produced by the op-amp here is fed directly back to one of its own inputs. Imagine if this device has an output that can have any value between −2V and +2V, but its inputs were only able to tolerate, or behave predictably with, inputs ranging from −1.5V to +1.5V. This op-amp could give itself a hernia.
In the case of your device, the MCP6V81, with supplies of \$V_{DD} = +2V\$ and \$V_{SS} = -2V\$, the output is able to reach almost +2V or almost −2V (within a few millivolts), and can swing around happily between those extremes.
For its inputs, the datasheet tells you that you can apply potentials in the range \$V_{SS}-0.2\$ to \$V_{DD}+0.3\$, which with supplies of ±2V would mean that any potential, between −2.2V and +2.3V, at either input, is acceptable.
By "acceptable" I mean that the device won't be damaged, and the function of the op-amp will be as you expect. The relationship between inputs (P and Q) and output (OUT) will conform closely to:
$$ V_{OUT} = A(V_P - V_Q) $$
where \$A\$ is some huge number representing the open-loop gain of the op-amp.
In other words, as long as your input voltages are within the allowable range −2.2V and +2.3V, and you don't do anything stupid with the output, that relationship will be obeyed. I won't go into what constitutes "stupid", except to say it includes anything connected to the output that prevents the op-amp from applying its own choice of potential there.
What happens when input potentials fall outside that acceptable range? Well, two things spring to mind. The first is that the op-amp may or may not have diodes in place which prevent you doing exactly that. Many devices have diodes inside the IC, like this:
If they are present, their primary purpose is to protect the whole op-amp from ESD, electrostatic discharge, but they have the side effect of also preventing you, the user, from explicitly applying potentials further than 0.3V or so beyond the power supply rails. Under such circumstances, they become highly conductive, bypassing current to the power supply rather than through their own sensitive circuitry.
If you do try to impose potentials significantly outside the power supply range, either so much current will flow that you destroy these diodes, swiftly followed by the destruction of everything else in the IC, or current demand will be so great that the source of the input signal is unable to maintain this over-voltage condition.
The other thing that can happen if inputs stray outside the acceptable range (assuming you don't damage anything in doing so), is that the op-amp's internal biasing is disturbed, and it enters a state in which it is no longer able able to promise that \$ V_{OUT} = A(V_P - V_Q) \$. Consequences vary from device to device. Older devices would suddenly swing their output from one extreme to the other, a phenomenon called "phase inversion", while modern devices tend to be better behaved.
As for your question, regarding the MCP6V81, with supplies of ±2V, you must never apply potentials to either input that are greater than +2.3V, or more negative than −2.2V. I don't know what will happen if you do, but I do know it will not be good.
Maybe you are asking "how is it possible to apply a potential higher than +2.3V, or lower than −2.2V, since the circuit's power supply is +2V and −2V?" Well, off course you can. Many systems have multiple supplies. Just because the supply you give this particular op-amp is ±2V, doesn't mean there aren't other sub-systems in the same circuit that use different supplies, and any one of them can produce signals that fall outside the range −2.2V to +2.3V.
And yes, if you are connecting the output of one such sub-system (with its own supplies) to an input of your op-amp, you'd better be sure that it's constrained to be within that op-amp's limits, otherwise at worst you break something, or at best you invoke some undefined op-amp behavior.
