Feedback systems & oscillations The transfer function of feedback system is:
$$
\frac{V_{\rm out}}{V_{\rm in}} = \frac{A}{1+Af}
$$
Where $A$ is the open loop gain, and $f$ is the feedback gain.

Now for oscillation to happen, $Af$ term (loop transmission) must be $-1$, which means the system is configured to be on a pole, because the denominator goes to zero. I have got couple of questions on this one. 


For oscillation to happen (steady state), the pole must be on $y$-axis (frequency axis) in the $s$-plane. I believe this is an assumption in the oscillation theory not reflected by the above condition that loop transmission is $-1$. Because loop transmission is $-1$ in all poles in the system. So it is up to the designer to make sure that the circuit operates on a pole on $y$-axis. Is this correct?
If I ignore all the Laplace transform thing, and just look at the transfer function 
$$\frac{V_{\rm out}}{V_{\rm in}} = \frac{A}{1+Af}$$ 
On the pole, the output is supposed to be very high (infinity). If I only consider this equation on standalone basis, then how come we have a steady state oscillation, the output voltage is supposed to be very high as denominator is zero. My explanation is that it does not happen, because the supply voltage to the Op-Amp/MOSFET (or whatever we use to implement the circuit) restricts that output voltage. Is this correct?
I think that the problem is that you are considering that $$ T = \frac{A}{1+Af} $$ is the actual transfer function of your system (an analog amplifier I guess) and this can be miss leading because that expression is just and approximation for the region of frequencys were the amplifier response is flat. If you want to make assumptions regarding the stability of the system you should work with the full Laplace transform, this is: $$ T(s) = \frac{A(s)}{1+A(s)f(s)} $$
In order to oscillate at a given frequency, the system needs to have conjugate poles on the imaginary-axis at that frequency, but in a practical case, if the system has poles on the right half-plane, as you said, the supply voltage of the amplifiers will move this poles into the j-axis

I don't have enough reputation to comment, but I think that this question that I asked a few weeks ago may be helpful.
https://electronics.stackexchange.com/questions/279365/positive-feedback-with-constant-transfer-functions