The difference in form of the term and counter-term in the renormalized Lagrangian [duplicate]

Consider, for example, the renormalized free field $\phi$-scalar Lagrangian (The question applies also to interacting $\phi^n$ theories, but that's not the issue):

$$\mathcal{L}=Z_2 \left(\partial_\mu\phi\right)^2 -Z_mm^2\phi^2=\mathcal{L}_0+\delta_2\left(\partial_\mu\phi\right)^2-\delta_m\phi^2$$

Where: $$\mathcal{L}_0=\left(\partial_\mu\phi\right)^2 -m^2\phi^2$$

The Feynman rules are:

My question is why are they different in form? Why the original term $\sim\frac{i}{p^2}$, yet the counter term $\sim p^2$.

Remark: The mass term can be disregraded in the answer, since I've established that by taking the mass as an interaction to a free-massless theory, we can take the massive-interaction $\sigma(p^2)$ as:

$$\sigma(p^2)=\frac{i}{p^2}+\frac{i}{p^2}\left(-im^2\right)\frac{i}{p^2}+\frac{i}{p^2}\left(-im^2\right)\frac{i}{p^2}\left(-im^2\right)\frac{i}{p^2}\dots=\frac{i}{p^2-m^2}$$

• @AccidentalFourierTransform, thank you for your answer. This almost answers my question. In your answer, you don't explain why the two forms are equivalent. According to what you wrote, I could be able to say that the original vertex goes like $ip^2$, but that's unphysical since I expect a pole in $p=0$. Why am I wrong? – EZLearner Aug 18 '17 at 11:32