Self energy, 1PI, and tadpoles

I'm going to give an explanation at the one loop level (which is the order of the diagrams given in the question).

At one loop, the effective action is given by $$ \Gamma[\phi]=S[\phi]+\frac{1}{2l}{\rm Tr}\log S^{(2)}[\phi],$$ where $S[\phi]$ is the classical (microscopic) action, $l$ is an ad hoc parameter introduced to count the loop order ($l$ is set to $1$ in the end), $S^{(n)}$ is the $n$th functional derivative with respect to $\phi$ and the trace is over momenta (and frequency if needed) as well as other indices (for the O(N) model, for example).

The physical value of the field $\bar\phi$ is defined such that $$\Gamma^{(1)}[\bar\phi]=0.$$ At the meanfield level ($O(l^0)$), $\bar\phi_0$ is the minimum of the classical action $S$, i.e. $$ S^{(1)}[\bar \phi_0]=0.$$ At one-loop, $\bar\phi=\bar\phi_0+\frac{1}{l}\bar\phi_1$ is such that $$S^{(1)}[\bar \phi]+\frac{1}{2l}{\rm Tr}\, S^{(3)}[\bar\phi].G_{c}[\bar\phi] =0,\;\;\;\;\;\;(1)$$ where $G_c[\phi]$ is the classical propagator, defined by $S^{(2)}[\phi].G_c[\phi]=1$. The dot corresponds to the matrix product (internal indices, momenta, etc.). The second term in $(1)$ corresponds to the tadpole diagram at one loop. Still to one-loop accuracy, $(1)$ is equivalent to $$ S^{(1)}[\bar \phi_0]+\frac{1}{l}\left(\bar\phi_1.\bar S^{(2)}+\frac{1}{2}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}\right)=0,\;\;\;\;\;\;(2) $$ where $\bar S^{(2)}\equiv S^{(2)}[\bar\phi_0] $, etc. We thus find $$\bar \phi_1=-\frac{1}{2}\bar G_c.{\rm Tr}\,\bar S^{(3)}.\bar G_c. \;\;\;\;\;\;(3)$$

Let's now compute the inverse propagator $\Gamma^{(2)}$. At a meanfield level, we have the meanfield propagator defined above $G_c[\bar\phi_0]=\bar G_c$ which is the inverse of $S^{(2)}[\bar\phi_0]=\bar S^{(2)}$. This is what is usually called the bare propagator $G_0$ in field theory, and is generalized here to broken symmetry phases.

What is the inverse propagator at one-loop ? It is given by $$\Gamma^{(2)}[\bar\phi]=S^{(2)}[\bar\phi]+\frac{1}{2l}{\rm Tr}\, \bar S^{(4)}.\bar G_{c}-\frac{1}{2l}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}. \bar S^{(3)}.\bar G_{c}, \;\;\;\;\;\;(4)$$ where we have already used the fact that the field can be set to $\bar\phi_0$ in the last two terms at one-loop accuracy. These two terms correspond to the first two diagrams in the OP's question. However, we are not done yet, and to be accurate at one-loop, we need to expand $S^{(2)}[\bar\phi]$ to order $1/l$, which gives $$\Gamma^{(2)}[\bar\phi]=\bar S^{(2)}+\frac{1}{l}\left(\bar S^{(3)}.\bar\phi_1+\frac{1}{2}{\rm Tr}\, \bar S^{(4)}.\bar G_{c}-\frac{1}{2}{\rm Tr}\, \bar S^{(3)}.\bar G_{c}. \bar S^{(3)}.\bar G_{c}\right). \;\;\;\;\;\;$$ Using equation $(3)$, we find $$\bar S^{(3)}.\bar\phi_1= -\frac{1}{2}\bar S^{(3)}.\bar G_c.{\rm Tr}\,\bar S^{(3)}.\bar G_c,$$ which corresponds to the third diagram of the OP. This is how these non-1PI diagrams get generated in the ordered phase, and they correspond to the renormalization of the order parameter (due to the fluctuations) in the classical propagator.

(I came across this question while looking through the "unanswered" category. Not sure why a 9-month-old question appeared in the list, but there still seems to be a room for my contribution.)

The quantum effective action $\Gamma[\phi]$ generates 1PI diagrams, and a tree-level analysis on it amounts to a full analysis on the original action $S[\phi]$.

The sum of all 2-point 1PI diagrams is what contributes (in addition to $p^{2}-m^{2}$, which is also present in the original action) to the quadratic part of $\Gamma[\phi]$. If there were no term linear in $\phi$ in $\Gamma[\phi]$, this would be the self energy itself; however, if $\Gamma[\phi]$ contained a linear term, one has to do an additional tree-level calculation to obtain the self energy from $\Gamma[\phi]$.

As for Feynman diagrams in the original post, the first two contribute to $\Gamma[\phi]$, and the third one appears when calculating the self energy from $\Gamma[\phi]$.