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
$$i_C(t) = -\frac{V}{\omega L} \sin \omega t$$
The integral $ \oint \vec{E} \cdot dl$ is zero for both the circuits.
Case 1
The integration path consists of the wire and the capacitor. Inside the capacitor there is a field, implying the integral is not zero and is the potential difference $V_0$. Hence for the total integral to be 0 , the integral over the wire is $-V_0$. Thus there is a potential difference between the ends of the wire and current flows.
Case 2
This time the entire wire is equivalent to the circuit. For any electric field the integral $ \int \vec{E} \cdot dl$ around a closed loop is 0 , and hence for a wire loop in an arbitrary electric field the integral or the potential difference between its ends is 0 implying no current.
The fringing field explanation is given in the text as a lot of people don't consider it and may incorrectly calculate the integral over the wire to be non zero.
Best Answer
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:
Hopefully, visualizing things this way can give you an intuitive grasp of how a capacitor and inductor work together to form a resonant circuit.