Renormalization conditions of the Callan-Symanzik equation Assume we have a massive $\phi^4$ theory the exact two-point correlation function is given as
$$G=\frac{iZ}{p^2-m_r^2}+\text{terms regular at }  p^2=m_r^2 $$ and if I want to apply renormalized perturbation theory I find
$$G=\frac{i}{p^2-m_r^2-\Sigma(p^2)}$$ 
where 
$$-i{\Sigma(p^2)}$$ is sum of all one particle irreducible diagrams.
Then the renormalization condition is 
$$-i{\Sigma(p^2)}|_{p^2=m_r^2}=0$$ such that it will have a pole with residue 1 at $p^2=m_r^2$.
However, when we derive Callan-Symanzik equation for massless theory we define the renormalization condition as
$$G=\frac{i}{p^2} \quad\text{at }\quad p^2=-M^2$$ where $M$ is renormalization scale. 
As far as I understood the first $G$ is exact, also the second one is exact and in order to match them we say  $$-i{\Sigma(p^2)}|_{p^2=m_r^2}=0$$.
But the confusing part is, in the massless case the exact $G$ should be exactly
$$G=\frac{iZ}{p^2}+\text{terms regular at }  p^2=0$$
and using renormalized perturbation theory we should find 
$$G=\frac{i}{p^2-\Sigma(p^2)}$$ 
then the renormalization condition is 
$$-i{\Sigma(p^2)}|_{p^2=0}=0$$ 
However, we use $p=M$ instead of $p=0$. This is confusing and I don't understand why. Also I don't understand whether in the massless case the exact two point correlation function is 
$$G=\frac{iZ}{p^2}+\text{terms regular at }  p^2=0$$ or not? If it is given like this and if we have also
$$G=\frac{i}{p^2} \quad\text{at }\quad p^2=-M^2$$
then are the terms $$(\text{terms regular at }  p^2=0 )=0\quad \text{ at } \quad p^2=-M^2  $$
I think the idea is they assume there is a momentum scale $M$ where theory behaves as a free theory exactly and that that point as renormalization condition  but how do they know if such a point exist.
 A: Massive scalar fields don't have singularities in their counterterms, however they have singularities when you take $m^2 \to 0$. For example, consider the counterterm $\delta_{\lambda}$ for the massive scalar field action
$$L = \frac{(\partial_{\mu} \phi_r)^2}{2} - \frac{(m \phi_r)^2}{2} - \frac{\lambda \phi_r^4}{4!} + \frac{\delta_Z (\partial_{\mu} \phi_r)^2}{2} - \frac{(\delta_m \phi_r)^2}{2} - \frac{\delta_{\lambda} \phi_r^4}{4!},$$
which we have obtained from the bare Lagrangian
$$L = \frac{(\partial_{\mu} \phi)^2}{2} - \frac{(m_0 \phi)^2}{2} - \frac{\lambda_0 \phi^4}{4!}$$
upon substituting 
$$\phi = Z^{1/2} \phi_r, \quad \delta_Z = Z-1, \quad\delta_m = m_0Z^2 - m^2, \quad \delta_{\lambda}  = \lambda_0 Z^2 - \lambda.$$
Hereby you can define the renormalization conditions as given in your question. One can now perform dimensional regularization and utilize the Feynmann parametrization to get an expression for $\delta_{\lambda}$ in $d=4$:
$$\delta_{\lambda} = \frac{\lambda^2}{32\pi^2} \int_0^1 dx \left( \frac{6}{\epsilon} - 3\gamma + 3 \log{4\pi } - \log{[m^2 - x(1-x)4m^2]} - 2\log{m^2}\right)$$
It is easy to see that this counterterm blows off when you take $m^2 \to 0$. As a result this renormalization scheme cannot be used when $m^2 \to 0$.
To avoid these singularities in the counterterms, you define the renormalization conditions for the massless case at some (unphysical) spacelike momentum $p$ with $p^2 = -M^2$, rather than on-shell $p^2 = 0$. The advantage is that the counterterms don't blow up as previously seen.
Note that this is generally a nice prescription (even for the massive case) as it completely bypasses the need to define renormalization conditions at the on-shell mass. One can now use various $n$-point Green functions so as to write down and solve the Callan-Symanzik equation, from which one derives the RG flow equation in terms of the renormalization scale $M$.   
