As the voltage between the capacitor's plates decreases, so should the
current flowing through the circuit.
I don't follow your reasoning here. Recall that, for an ideal capacitor, we have:
$$i_C = C\frac{dv_C}{dt}$$
In words, the current through the capacitor is proportional to the rate of change of the voltage across, not the instantaneous value of the voltage.
So, for example, if the voltage across the capacitor is sinusoidal
$$v_C = V \sin\omega t$$
the current is
$$i_C = \omega CV \cos \omega t$$
which means (1) that the maximum current (magnitude) occurs when the voltage is zero and (2) that the maximum voltage (magnitude) occurs when the current is zero.
Now, for this simple LC circuit, the voltage across the capacitor is identical to the voltage across the inductor:
$$v_C = v_L$$
thus,
$$i_C = C\frac{dv_L}{dt}$$
For an ideal inductor, we have:
$$v_L = L\frac{di_L}{dt}$$
But, the inductor current is
$$i_L = - i_C$$
thus,
$$i_C = -LC\dfrac{d^2i_C}{dt^2}$$
which means that the current is sinusoidal
$$i_C = A \sin \omega t + B \cos \omega t $$
where
$$\omega = \frac{1}{\sqrt{LC}}$$
Since, in your example, the initial current is zero and the initial voltage is $V$, we have
$$i_C(t) = -\frac{V}{\omega L} \sin \omega t$$
Wouldn't this inductor's emf counteract the discharging capacitor and actually charge it? / stop the capacitor from fully discharging?
The inductor doesn't care about what the charge state of the capacitor is. All it cares about is how quickly the current through it is changing, and it generates a back-voltage according to the equation V=L*dI/dt. You can think of an inductor as giving "momentum" to the current. If the current is zero, then it wants to keep the current zero. If the current is non-zero, it wants to keep the current at that same non-zero value. If the current is increasing, it generates a counter-voltage acting in the opposite direction to the current flow.
The analogy I like to use is a circuit of water pipes in which inductors are represented by a heavy propellor in a water pipe. If water flow is suddenly turned on, the heavy propellor initially resists the flow of water. But over time the propellor spins faster in response to the water flow. If the water flow past the propellor is then reduced, the heavy propellor resists the decrease in water flow because it is now spinning fast and tries to continue pushing the water through the pipe. This is analogous to how an inductor resists changes in the electrical current flow through it.
Using this water circuit analogy, a capacitor can be represented as a section of pipe which has a rubber membrane stretched across the inside of it. If you push water into one end of this pipe section, the rubber membrane stretches and creates a back pressure resisting attempts to push more and more water into it. If you then stop applying water pressure to that side of the pipe section, the rubber membrane springs back to its flat, equilibrium position, pushing the water back out the same side of the pipe as you were trying to push the water in. This is analogous to how a capacitor "pushes back" with a back-voltage when you push electrical charge into a capacitor.
If you make a closed electrical circuit with this heavy propellor (which represents the inductor) and the rubber-membrane pipe section (which represents the capacitor), then you should be able to see how a resonant water oscillation in the circuit can be set up. Imagine the rubber membrane pipe section being "charged" by forcing water into one side. When you release the applied pressure, water will flow past the heavy propellor, which will then speed up and try to maintain a constant water flow past it. However, as the water flows past the heavy propellor and into the other side of the rubber membrane pipe section, the rubber membrane goes to its equilibrium position and then starts getting stretched in the opposite direction. Eventually, the back-pressure becomes so large that the direction of water flow is reversed and the cycle happens all over again.
In summary, with this analogy we have the following:
electrical current <-> water flow
voltage <-> water pressure
inductor <-> heavy propellor
capacitor <-> rubber membrane pipe section
Hopefully, visualizing things this way can give you an intuitive grasp of how a capacitor and inductor work together to form a resonant circuit.
Best Answer
It's true that the initially clock-wise current that discharges the capacitor is increasing at first and the reason for this is that the voltage across the inductor is initially positive (the top terminal of the inductor is more positive than the bottom terminal).
However, and in contrast to your statement quoted above, the current will continue to increase even as the voltage across the capacitor (and the inductor in parallel) decreases towards zero. As long as the voltage is positive, the current must be increasing. Why? It follows from the fundamental inductor equation:
$$v_L(t) = L \frac{di_L}{dt}$$
Since $L$ is a positive constant, a positive inductor voltage requires a positive rate of change of inductor current - positive voltage across implies increasing current through.
When the capacitor is fully discharged, the voltage across is zero and the current is maximum (the rate of change of current is zero at the maximum) but now the current begins to charge the capacitor 'in the opposite direction' and the voltage across becomes negative which implies a negative rate of change of inductor current.
In short, the current doesn't begin to decrease until the voltage across passes through zero and becomes negative.