Why does the counterterm's propagator have inverse units of the propagator? $\phi^4$-theory According to Peskin & Schroeder (page 325), the Feynman rule for the counterterm
 ------(x)----- 

for 
$$ \frac12 \delta_Z(\partial_\mu\phi_r)^2-\frac12\delta_m \phi_r^2$$
being $\phi_r$ the renormalized field, is given by 
$$i(p^2\delta_Z-\delta_m)$$
which resembles rather the  (multiplicative) inverse of the propagator for the original Lagrangian (whith physical quantities). Why?
 A: Thee diverging terms for the propagator come from the renormalization of the self-energy $\Sigma$, defined by $G^{-1}=G_0^{-1}-\Sigma$, where $G_0$ is the propagator defined by the Lagrangian (i.e. bare propagator + counterterms) :
$G^{-1}_0=(1+\delta Z)p^2+(m^2_0+\delta m^2)$.
One chooses the counterterms to cancel the divergences coming from $\Sigma$ order by order. If the theory is perturbatively renormalizable, only these two counterterms are sufficient at every order in perturbation theory. 
A: Counter Term Feynman Rules
A counter term in a Lagrangian is interpreted as an interaction, and the Feynman rules may be derived by the Feynman path integral in the standard way. This accounts for the factor of $i$ and the functional derivatives will take care of the factor of $\frac12$.
Just like in $\frac{\lambda}{4!}\phi^4$ theory we have a $-i\lambda$ Feynman rule, with $\frac{\delta_m}{2}\phi^2$ we have a $-i\delta_m$ Feynman rule. In addition, derivatives of fields give rise to momenta: $(\partial \phi)^2 \mapsto p^2$. So the overall counter term is,
$$i(p^2\delta_Z-\delta_m).$$
The key is, if in doubt, always check things explicitly by treating the terms as interactions in the Feynman path integral. 

Alternative Method
I highly recommend Collins' book on renormalisation for a method of renormalisation which dispenses with a Lagrangian and is ideal for systematically computing diagrams. This involves a combination of the BPHZ method and dimensional regularisation.
Essentially, the Zimmermann forest formula generates all the counter term graphs. Then instead of inserting Feynman rules, you insert a particular 'subtraction' operator to remove the divergences.
Of course, this can be related back to a Lagrangian by considering what the Feynman rules would read, and equating the two results to find the counter terms to whatever order you need.
A: I also found it hard to find a rigorous explanation. By dimensional analysis and explicit checking you indeed find that this is the correct Feynman rule for the counterterm. 
The closest thing I have to a derivation is that we can include the counterterm in the kinetic term and expand this full propagator for small $\delta Z$
$ G = \frac{i}{(1+\delta Z)p^2+m^2} \approx \frac{1}{p^2+m^2}-\frac{\delta Z}{(p^2+m^2)^2} + ...$. 
I think that not including the $(p^2+m^2)^2$ term in the Feynman rule is correct since this counterterm gets treated as an (amputated) vertex. A four point function similarly does not include 4 propagator factors in its feynman rule.
I do not know why I get a different sign than Peskin & Schroeder.
Also, the expansion is questionable because $\delta Z$ is in fact infinite as opposed to small.
