Doubt about Poles of the Transfer Function
For a system to be stable, the poles of the TF must be in the LHP. RHP can have poles if the system is not stable.

I don't remember the textbook proof of it. But the way I understood is explained below.

Consider a transfer function with two poles. One at RHP (\$p_1\$) and the other at LHP (\$-p_2\$). Then the transfer function will be of the form:
$$\frac{Y(s)}{X(s)} = \frac{A}{(s-p_1)(s+p_2)}$$

Assuming unit step input, \$x(t)= u(t),\$  output will be
$$Y(s) = \frac{A}{(s-p_1)(s+p_2)}\times \frac{1}{s}$$

Using partial fraction, Y(s) can be written as,
$$Y(s) = \frac{A_1}{s-p_1} + \frac{A_2}{s+p_2} + \frac{A_3}{s}$$

Where \$A_1, A_2, A_3\$ are constants. 
Taking Laplace inverse,

$$y(t) = (A_1e^{p_1} + A_2e^{-p_2} + A_3)u(t)$$

The first term \$A_1e^{p_1}\$ will go unbounded as \$t\rightarrow\infty\$ and so does the output. But the input is still bounded. So the system is not stable (see BIBO stability).

The reason for instability is the presence of \$A_1e^{p_1}\$ term in the output which is contributed by the pole at RHP.

Conclusion: The pole at RHP can contribute to an exponentially growing term in the output of the system which will lead to instability.