Yeah no problem. Lets name the inverters from left to right: inv1, inv2, inv3 and so on.
We will look at the top circuit first.
If this was a physical circuit you would not need an initial input, as the circuit would self-initialize. However for the sake of completeness lets say the input at t=0 is 1. After t=0 the input is not driven. Follow the signal down the chain.
At t=050ns the output of inv1 is 0
At t=100ns the output of inv2 is 1
At t=150ns the output of inv3 is 0
At t=200ns the output of inv4 is 1
At t=250ns the output of inv5 is 0
Now the output of the ring oscillator has stabilized at 0. This is the same signal as the input to inv1. The input to inv1 was initially 1, but now it is 0. Therefore,
At t=300ns the output of inv1 is 1
At t=350ns the output of inv2 is 0
At t=400ns the output of inv3 is 1
At t=450ns the output of inv4 is 0
At t=500ns the output of inv5 is 1
An observation as to the period of the ring oscillator can now be made. At t=250ns the output of inv5 was 0 and it changed to 1 at t=500ns. Therefore since 250ns is half of the period, 500ns is the full period. In fact, the formula for the period for an arbitrary ring oscillator will be 2x(number of inverters)x(inverter delay).
Now, lets look at the case of 6 inverters. Again say the input at t=0 is 1. After t=0 the input is not driven.
At t=050ns the output of inv1 is 0
At t=100ns the output of inv2 is 1
At t=150ns the output of inv3 is 0
At t=200ns the output of inv4 is 1
At t=250ns the output of inv5 is 0
At t=300ns the output of inv6 is 1
At t=350ns the output of inv1 is still 0
A ring oscillator must have an odd number of inverters. If you have a chain of an even number of inverters it will behave rather like a D Flip-Flop. Can I get some points now?