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.
Weinberg, QFT 2, in Section 16.1 in a footnote 2 refers to Coleman, Aspects of Symmetry, p. 135-6, which features the $\hbar$/loop expansion. See also Refs. 3 & 4 for a similar idea. In this answer we provide a non-inductive argument along these lines. A nice feature of this argument is that we do not have to deal explicitly with pesky combinatorics and symmetry factors of individual Feynman diagrams. This is already hardwired into the formalism.
A) Let us first recall some basic facts from field theory. The classical (=$\hbar$-independent) action
$$S[\phi]~\equiv~\underbrace{\frac{1}{2}\phi^k (S_2)_{k\ell}\phi^{\ell}}_{\text{quadratic part}} + \underbrace{S_{\neq 2}[\phi]}_{\text{the rest}}, \tag{A1}$$
is the generating functional for bare vertices (and inverse bare propagator $(S_2)_{k\ell}$).
The partition function/path integral is
$$\begin{align} Z[J] ~:=~&\int \! {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar} \underbrace{\left(S[\phi]+J_k \phi^k\right)}_{=:~S_J[\phi]}\right\}\tag{A2} \cr
~\stackrel{\begin{array}{c}\text{Gauss.}\cr\text{int.}\end{array}}{\sim}&
{\rm Det}\left(\frac{1}{i} (S_2)_{mn}\right)^{-1/2}\cr
&\exp\left\{\frac{i}{\hbar} S_{\neq 2}\left[ \frac{\hbar}{i} \frac{\delta}{\delta J}\right] \right\}\cr
&\exp\left\{- \frac{i}{2\hbar} J_k (S_2^{-1})^{k\ell} J_{\ell} \right\}\tag{A3}\cr
~\stackrel{\begin{array}{c}\text{WKB}\cr\text{approx.}\end{array}}{\sim}&
{\rm Det}\left(\frac{1}{i}\frac{\delta^2 S[\phi[J]]}{\delta \phi^m \delta \phi^n}\right)^{-1/2}\cr
&\exp\left\{ \frac{i}{\hbar}\left(S[\phi[J]]+J_k \phi^k[J]\right)\right\}\cr
&\left(1+ {\cal O}(\hbar)\right) \tag{A4}\end{align} $$
in the stationary phase/WKB approximation $\hbar\to 0$. In eq. (A4)
$$ J_k~\approx~-\frac{\delta S[\phi]}{\delta \phi^k} \qquad \Leftrightarrow \qquad \phi^k~\approx~\phi^k[J] \tag{A5}$$
are the Euler-Lagrange (EL) equations for the quantum field $\phi^k$.
Notice in the diagram expansion (A3) how a bare vertex comes with $\hbar$-weight $=-1$; an internal bare propagator $(S_2^{-1})^{k\ell}$ comes with $\hbar$-weight $=+1$; and an external leg comes with $\hbar$-weight $=0$.
The linked cluster theorem states that the generating functional for connected diagrams is
$$ W_c[J]~=~\frac{\hbar}{i}\ln Z[J], \tag{A6}$$
cf. e.g. this Phys.SE post. Note that the connected vacuum bubbles $W_c[J\!=\!0]=\frac{\hbar}{i}\ln Z[J\!=\!0]$ by definition is correlated to the normalization of the path integral, and hence is not physically relevant. (We allow the possibility that it is non-zero to be as general as possible.)
Next recall the $\hbar$/loop-expansion
$$ L~=~I-V+1, \tag{A7} $$
cf. my Phys.SE answer here.
The $\hbar$/loop-expansion together with eqs. (A4) & (A6) imply that the generating functional
$$ W_{c}^{\rm tree}[J]~\stackrel{(A4)+(A6)}{=}~S[\phi] + J_i \phi^i \tag{A8}$$
for connected tree diagrams is the Legendre transformation of the classical action. Note that the EL eqs. (A5) are compatible with this.
Eqs. (A3) and (A6) yield
$$\begin{align} &W^{\rm tree}_c[J]~\stackrel{(A3)+(A6)}{=}\cr \lim_{\hbar\to 0} \frac{\hbar}{i}& \ln\left( \exp\left\{ \frac{i}{\hbar} S_{\neq 2}\left[ \frac{\hbar}{i} \frac{\delta}{\delta J}\right] \right\}
\exp\left\{- \frac{i}{2\hbar} J_k (S_2^{-1})^{k\ell} J_{\ell} \right\} \right). \end{align}\tag{A9}$$
Notice how eq. (A9) only refers to objects in eqs. (A1) & (A8), and hence can be viewed as a consequence of them alone.
Eq. (A9) realizes the fact that given an arbitrary finite set of external source insertions, then (a sum of all possible) connected tree diagrams is (a sum of all possible) trees of bare propagators $(S_2^{-1})^{k\ell}$ and bare vertices.
Note that the one-loop square root factors in eqs. (A3) & (A4) do not affect the zero-loop/tree formula (A9) & (A8), respectively.
$\downarrow$ Table 1: Structural similarity between Sections A & B.
$$ \begin{array}{ccc} A &\leftrightarrow & B \cr
\phi^k&\leftrightarrow & \phi_{\rm cl}^k \cr
S[\phi]&\leftrightarrow &\Gamma[\phi_{\rm cl}]\cr
\hbar&\leftrightarrow &\hbar^{\prime} \cr
Z[J]&\leftrightarrow &Z_{\Gamma}[J]\cr
W^{\rm tree}_c[J]&\leftrightarrow &W_c[J]
\end{array}$$
B) Finally let us address OP's question. Consider the effective/proper action
$$ \Gamma[\phi_{\rm cl}]~\equiv~\underbrace{\frac{1}{2}\phi_{\rm cl}^k (\Gamma_2)_{k\ell}\phi_{\rm cl}^{\ell}}_{\text{quadratic part}} + \underbrace{\Gamma_{\neq 2}[\phi_{\rm cl}]}_{\text{the rest}}.\tag{B1}$$
Unlike the classical action (A1), the effective action (B1) depends (implicitly) on Planck's reduced constant $\hbar$. We would like to make a loop-expansion wrt. a new parameter $\hbar^{\prime}$.
To this end, define a partition function/path integral
$$\begin{align} Z_{\Gamma}[J] ~:=~&\int \! {\cal D}\frac{\phi_{\rm cl}}{\sqrt{\hbar^{\prime}}}~\exp\left\{ \frac{i}{\hbar^{\prime}} \underbrace{\left(\Gamma[\phi_{\rm cl}]+J_k \phi_{\rm cl}^k\right)}_{=:~\Gamma_J[\phi_{\rm cl}]}\right\} \tag{B2}\cr
~\stackrel{\begin{array}{c}\text{Gauss.}\cr\text{int.}\end{array}}{\sim}&
{\rm Det}\left(\frac{1}{i} (\Gamma_2)_{mn}\right)^{-1/2}\cr
&\exp\left\{ \frac{i}{\hbar^{\prime}} \Gamma_{\neq 2}\left[ \frac{\hbar^{\prime}}{i} \frac{\delta}{\delta J}\right] \right\}\cr
&\exp\left\{- \frac{i}{2\hbar^{\prime}} J_k (\Gamma_2^{-1})^{k\ell} J_{\ell} \right\}\tag{B3}\cr
~\stackrel{\begin{array}{c}\text{WKB}\cr\text{approx.}\end{array}}{\sim}&
{\rm Det}\left(\frac{1}{i}\frac{\delta^2\Gamma[\phi_{\rm cl}[J]]}{\delta \phi_{\rm cl}^m \delta \phi_{\rm cl}^n}\right)^{-1/2}\cr
&\exp\left\{ \frac{i}{\hbar^{\prime}}\left(\Gamma[\phi_{\rm cl}[J]]+J_k \phi_{\rm cl}^k[J]\right)\right\}\cr
&\left(1+ {\cal O}(\hbar^{\prime})\right) \tag{B4}\end{align} $$
in the stationary phase/WKB approximation $\hbar^{\prime}\to 0$. Also the EL eqs. for the effective action $\Gamma_J[\phi_{\rm cl}]$ for the classical field $\phi_{\rm cl}^k$ read
$$ J_k~\approx~-\frac{\delta \Gamma[\phi_{\rm cl}]}{\delta \phi_{\rm cl}^k} \qquad \Leftrightarrow \qquad \phi_{\rm cl}^k~\approx~\phi_{\rm cl}^k[J]. \tag{B5}$$
Recall that the effective action (B1) is by definition the Legendre transformation of the generating functional
$$ W_{c}[J]~\equiv~\Gamma[\phi_{\rm cl}] + J_k \phi_{\rm cl}^k \tag{B8}$$
for connected diagrams. Note that the EL eqs. (B5) are compatible with this.
Due to the structural similarity between two the Legendre transformations (A8) & (B8), cf. Table 1, we obtain an analogue to eq. (A9):
$$\begin{align}&W_c[J]~\stackrel{(B3)+(B4)+(B8)}{=}\cr \lim_{\hbar^{\prime}\to 0} \frac{\hbar^{\prime}}{i}& \ln\left( \exp\left\{ \frac{i}{\hbar^{\prime}} \Gamma_{\neq 2}\left[ \frac{\hbar^{\prime}}{i} \frac{\delta}{\delta J}\right] \right\} \exp\left\{- \frac{i}{2\hbar^{\prime}} J_k (\Gamma_2^{-1})^{k\ell} J_{\ell} \right\} \right) .\end{align}\tag{B9}$$
In retrospect, eq. (B9) can be viewed as a functorial consequence of eqs. (B1) & (B8) alone.
On the other hand, given an arbitrary finite set of external source insertions, then (a sum of all possible) connected diagrams is (a sum of all possible) trees of full propagators$^{\dagger}$ $(\Gamma_2^{-1})^{k\ell}$ and (amputated) 1PI vertices, cf. Lemma 3.11 in Ref. 5.
Together with eq. (B9), we conclude that the effective action $\Gamma[\phi_{\rm cl}]$ is the generating functional for (amputated) 1PI vertices (and inverse full propagator $(\Gamma_2)_{k\ell}$). $\Box$
References:
S. Weinberg, Quantum Theory of Fields, Vol. 2, 1995; Section 16.1.
S. Coleman, Aspects of Symmetry, 1985; p. 135-6.
M. Srednicki, QFT, 2007; Chapter 21. A prepublication draft PDF file is available here.
D. Skinner, QFT in 0D, p. 32. (Hat tip: The Last Knight of Silk Road.)
P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Sections 3.11 & 3.12. (Hat tip: Abdelmalek Abdesselam.)
R. Kleiss, Pictures, Paths, Particles, Processes, Feynman Diagrams and All That and the Standard Model, lecture notes, 2013; section 1.5.2.
--
$^{\dagger}$ Fine print:
Assume that the generator $W_c[J]$ of connected diagrams has no terms linear in $J$, so that the effective action $\Gamma[\phi_{\rm cl}]$ has no terms linear in $\phi_{\rm cl}$, and so that $(\Gamma_2^{-1})^{k\ell}=-(W_{c,2})^{k\ell}$ is the full connected propagator, cf. my Phys.SE answer here.
Here the notion of the one-particle irreducible (1PI) vertices are defined wrt. to full propagators $(W_{c,2})^{k\ell}$, which is equivalent to the notion of 1PI vertices wrt. to bare propagators $(S_2^{-1})^{k\ell}$, cf. e.g. this Phys.SE post.
Best Answer
The object $\Gamma[\phi]$ (I'll drop the subscript to keep notation more simple) is a real number. We'd like to expand it in a Taylor series with respect to its dependence on the field variable $\phi$. If this was a regular function, we'd have $$\Gamma[\phi] = \sum_{n=0}^{+\infty} \frac{1}{n!} \frac{\partial^n \Gamma}{\partial \phi^n} \phi^n.$$
However, $\phi$ has a dependence on spacetime. When we differentiate with respect to $\phi$, we must take that into consideration. This amounts to $$\Gamma[\phi] = \sum_{n=0}^{+\infty} \frac{1}{n!} \int \frac{\delta^n \Gamma}{\delta\phi(x_1) \cdots \delta\phi(x_n)} \phi(x_1) \cdots \phi(x_n) \textrm{d}^d{x_1} \cdots \textrm{d}^d{x_n},$$ where each term is evaluated at $\phi = 0$.
To see this, let us pick an example. Consider $\Gamma[\phi] = \frac{m^2}{2} \int \phi^2 \textrm{d}^d{x}$. Using P&S's Eq. (9.31), we have $$\frac{\delta \Gamma}{\delta \phi(x)} = m^2 \phi(x)$$ and $$\frac{\delta^2 \Gamma}{\delta \phi(y) \delta \phi(x)} = m^2 \delta^{(d)}(x - y).$$ The remaining derivatives will vanish. If we plug in these expressions on the functional Taylor series I gave above, we have $$\Gamma[\phi] = \frac{m^2}{2} \int 0 \ \textrm{d}^d{x} + m^2 \int 0 \cdot \phi(x) \textrm{d}^d{x} + \frac{1}{2} m^2 \int \phi(x)\phi(y) \delta^{(d)}(x - y) \textrm{d}^d{x}\textrm{d}^d{y} = \frac{m^2}{2} \int \phi^2 \textrm{d}^d{x},$$ which recovers exatcly the original expression.
To reach P&S's Eq. (13.20), we still have to employ Eq. (11.96). From the functional Taylor series expression, it leads to $$\Gamma[\phi] = i \sum_{n=0}^{+\infty} \frac{1}{n!} \int \langle\hat{\phi}(x_1) \cdots \hat{\phi}(x_n) \rangle_{\text{1PI}} \phi(x_1) \cdots \phi(x_n) \textrm{d}^d{x_1} \cdots \textrm{d}^d{x_n},$$ where I use hats to denote the quantum fields. On Eq. (13.20), P&S change notation according to $\Gamma^{(n)}(x_1, \ldots, x_n) = \langle\hat{\phi}(x_1) \cdots \hat{\phi}(x_n) \rangle_{\text{1PI}}$, which leads to
$$\Gamma[\phi] = i \sum_{n=0}^{+\infty} \frac{1}{n!} \int \Gamma^{(n)}(x_1, \ldots, x_n) \phi(x_1) \cdots \phi(x_n) \textrm{d}^d{x_1} \cdots \textrm{d}^d{x_n},$$ as desired.
Functional derivatives occur not only on QFT, but also on Classical Field Theory. I suggest Section 11.3 of Nivaldo Lemos' Analytical Mechanics. It is an excellent, mathematically clear book with a nice account of Classical Field Theory. While it is not a Math book, it is much more careful than most Physics books and quite a good companion for QFT (and for life, if you ask me).