Capacitor Discharging through a fixed resistor

As mentioned in a comment, the capacitor acts as a source of charge when the switch is set to position 2. Charge from the capacitor will flow as a transient current in the circuit which declines exponentially to zero. The current flows through the resistor. At any given moment the resistor will have an instantaneous potential difference that will fall off in proportion to the current.

Roughly, V = iR, so the profile of voltage across the resistor should look like the profile for the capacitor.

Remember that (roughly) the electrons are leaving one side of the plate and flowing through the resistor. The potential difference across the capacitor is of opposite sign to that of the resistor, so sum of the voltages is zero, as you note.

but as these two are the only components in the circuit, then we should have Vcapacitor + Vresistor=0

This is true or not depending on the voltage reference polarity chosen for the capacitor and resistor.

For the circuit consisting of the series connected source, capacitor, and resistor, one would typically choose to label the left-most terminal of the capacitor as positive and the top-most terminal of the resistor as positive.

This is physically equivalent to choosing to place the red lead of your voltmeter on the left-most terminal of the capacitor and the black lead on the right-most terminal and similarly for the resistor.

In that case, KVL yields

$$Delta V = v_C + v_R$$

When the switch is thrown from position 1 to position 2, KVL yields

$$0 = v_C + v_R$$

and so the voltage across the resistor is the opposite sign (and same magnitude) as the voltage across the capacitor.

But this is merely a convention. Changing the voltage reference polarity of either the capacitor or the resistor (equivalent to swapping your voltmeter leads around) would give $v_C = v_R$ when the switch is in position 2.