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AccidentalFourierTransform's above comment is exactly right: The point is that Srednicki's time-ordering $T$ should be replaced with covariant time-ordering $T_{\rm cov}$, i.e. time-differentiations inside its argument should be taken after/outside the usual time ordering $T$.

This resolves the apparent conflict/contradiction between Srednicki's eqs. (22.23') and (22.24). In other words, the 2-pt function $\left<T\{\phi\phi\} \right>$ can only be the Greens function $\Delta$ if we use $T_{\rm cov}$ instead of $T$ in the SD eq. (22.23).

More generally, the formal correspondence/dictionary between $$\text{operator formulation} \quad\leftrightarrow\quad \text{path integral formulation}\tag{A} $$ is $$\begin{align} \left< \Omega \left| T_{\rm cov}\{ F[\phi]\} \right| \Omega \right>_J ~=~& \frac{1}{Z[J]}\int \! {\cal D}\phi~F[\phi]~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\}\cr ~=~& \frac{1}{Z[J]}F\left[\frac{\hbar}{i}\frac{\delta}{\delta J} \right]Z[J],\end{align}\tag{B} $$ where $F$ is an arbitrary functional and where the partition function/path integral is $$\begin{align} Z[J]~:=~& \int \! {\cal D}\phi~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\},\cr S[\phi;J]~:=~&S[\phi]+J_k\phi^k, \end{align}\tag{C}$$ The correspondence (B) follows from the underlying time slicing procedure of path integrals. See e.g. this and this Phys.SE answer.

Now to the main point: Notice how the dictionary (B) naturally talks to $T_{\rm cov}$ rather than $T$: If the functional $F$ doesn't contain time derivatives, it doesn't matter whether we use $T$ or $T_{\rm cov}$. However, if $F$ does contain time derivatives, they get applied outside the correlator, i.e. the time-order is $T_{\rm cov}$.

Then the Schwinger-Dyson (SD) equations becomes $$ \left< \Omega \left| T_{\rm cov}\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~i\hbar\left< \Omega \left| T_{\rm cov}\left\{\frac{\delta F[\phi]}{\delta \phi(x)} \right\}\right| \Omega \right>_J ~, \tag{D}$$ cf. e.g. this Phys.SE post.

In contrast, if we only use the usual time ordering $T$, we do not get the contact term: $$ \left< \Omega \left| T\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~0, \tag{E}$$ because the EOMs are satisfied in quantum average, cf. e.g. this Phys.SE post.

AccidentalFourierTransform's above comment is exactly right: The point is that Srednicki's time-ordering $T$ should be replaced with covariant time-ordering $T_{\rm cov}$, i.e. time-differentiations inside its argument should be taken after/outside the usual time ordering $T$.

This resolves the apparent conflict/contradiction between Srednicki's eqs. (22.23') and (22.24). In other words, the 2-pt function $\left<T\{\phi\phi\} \right>$ can only be the Greens function $\Delta$ if we use $T_{\rm cov}$ instead of $T$ in the SD eq. (22.23).

More generally, the formal correspondence/dictionary between $$\text{operator formulation} \quad\leftrightarrow\quad \text{path integral formulation}\tag{A} $$ is $$\begin{align} \left< \Omega \left| T_{\rm cov}\{ F[\phi]\} \right| \Omega \right>_J ~=~& \frac{1}{Z[J]}\int \! {\cal D}\phi~F[\phi]~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\}\cr ~=~& \frac{1}{Z[J]}F\left[\frac{\hbar}{i}\frac{\delta}{\delta J} \right]Z[J],\end{align}\tag{B} $$ where $F$ is an arbitrary functional and where the partition function/path integral is $$\begin{align} Z[J]~:=~& \int \! {\cal D}\phi~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\},\cr S[\phi;J]~:=~&S[\phi]+J_k\phi^k, \end{align}\tag{C}$$ The correspondence (B) follows from the underlying time slicing procedure of path integrals. See e.g. this and this Phys.SE answer.

Now to the main point: Notice how the dictionary (B) naturally talks to $T_{\rm cov}$ rather than $T$: If the functional $F$ doesn't contain time derivatives, it doesn't matter whether we use $T$ or $T_{\rm cov}$. However, if $F$ does contain time derivatives, they get applied outside the correlator, i.e. the time-order is $T_{\rm cov}$.

Example. If $$ F[\phi]~=~\prod_{i=1}^n \left(\frac{\partial}{\partial t_i} \right)^{m_i} \phi(t_i),\tag{D}$$ then $$\begin{align}\frac{1}{Z[J]}&F\left[\frac{\hbar}{i}\frac{\delta}{\delta J} \right]Z[J]\cr ~\stackrel{(D)}{=}~&\frac{1}{Z[J]}\left[\prod_{i=1}^n \left(\frac{\partial}{\partial t_i} \right)^{m_i}\right]\int \! {\cal D}\phi~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\}\prod_{j=1}^n\phi(t_j)\cr ~\stackrel{(B)}{=}~&\left[\prod_{i=1}^n \left(\frac{\partial}{\partial t_i} \right)^{m_i}\right]\left< \Omega \left| T\left\{ \prod_{j=1}^n\phi(t_j)\right\} \right| \Omega \right>_J\cr ~\stackrel{(D)}{=}~&\left< \Omega \left| T_{\rm cov}\{ F[\phi]\} \right| \Omega \right>_J. \end{align}\tag{E}$$

Then the Schwinger-Dyson (SD) equations becomes $$ \left< \Omega \left| T_{\rm cov}\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~i\hbar\left< \Omega \left| T_{\rm cov}\left\{\frac{\delta F[\phi]}{\delta \phi(x)} \right\}\right| \Omega \right>_J ~, \tag{F}$$ cf. e.g. this Phys.SE post.

In contrast, if we only use the usual time ordering $T$, we do not get the contact term: $$ \left< \Omega \left| T\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~0, \tag{G}$$ because the EOMs are satisfied in quantum average, cf. e.g. this Phys.SE post.

AccidentalFourierTransform's above comment is exactly right: The point is that Srednicki's time-ordering $T$ should be replaced with covariant time-ordering $T_{\rm cov}$, i.e. time-differentiations inside its argument should be taken after/outside the usual time ordering $T$.

This resolves the apparent conflict/contradiction between Srednicki's eqs. (22.23') and (22.24). In other words, the 2-pt function $\left<T\{\phi\phi\} \right>$ can only be the Greens function $\Delta$ if we use $T_{\rm cov}$ instead of $T$ in the SD eq. (22.23).

More generally, the formal correspondence/dictionary between $$\text{operator formulation} \quad\leftrightarrow\quad \text{path integral formulation}\tag{A} $$ is $$\begin{align} \left< \Omega \left| T_{\rm cov}\{ F[\phi]\} \right| \Omega \right>_J ~=~& \frac{1}{Z[J]}\int \! {\cal D}\phi~F[\phi]~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\}\cr ~=~& \frac{1}{Z[J]}F\left[\frac{\hbar}{i}\frac{\delta}{\delta J} \right]Z[J],\end{align}\tag{B} $$ where $F$ is an arbitrary functional and where the partition function/path integral is $$\begin{align} Z[J]~:=~& \int \! {\cal D}\phi~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\},\cr S[\phi;J]~:=~&S[\phi]+J_k\phi^k, \end{align}\tag{C}$$ The correspondence (B) follows from the underlying time slicing procedure of path integrals. See e.g. this and this Phys.SE answer. Notice how the dictionary (B) naturally talks to $T_{\rm cov}$ rather than $T$.

Then the Schwinger-Dyson (SD) equations becomes $$ \left< \Omega \left| T_{\rm cov}\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~i\hbar\left< \Omega \left| T_{\rm cov}\left\{\frac{\delta F[\phi]}{\delta \phi(x)} \right\}\right| \Omega \right>_J ~, \tag{D}$$ cf. e.g. this Phys.SE post.

In contrast, if we only use the usual time ordering $T$, we do not get the contact term: $$ \left< \Omega \left| T\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~0, \tag{E}$$ because the EOMs are satisfied in quantum average, cf. e.g. this Phys.SE post.

AccidentalFourierTransform's above comment is exactly right: The point is that Srednicki's time-ordering $T$ should be replaced with covariant time-ordering $T_{\rm cov}$, i.e. time-differentiations inside its argument should be taken after/outside the usual time ordering $T$.

This resolves the apparent conflict/contradiction between Srednicki's eqs. (22.23') and (22.24). In other words, the 2-pt function $\left<T\{\phi\phi\} \right>$ can only be the Greens function $\Delta$ if we use $T_{\rm cov}$ instead of $T$ in the SD eq. (22.23).

More generally, the formal correspondence/dictionary between $$\text{operator formulation} \quad\leftrightarrow\quad \text{path integral formulation}\tag{A} $$ is $$\begin{align} \left< \Omega \left| T_{\rm cov}\{ F[\phi]\} \right| \Omega \right>_J ~=~& \frac{1}{Z[J]}\int \! {\cal D}\phi~F[\phi]~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\}\cr ~=~& \frac{1}{Z[J]}F\left[\frac{\hbar}{i}\frac{\delta}{\delta J} \right]Z[J],\end{align}\tag{B} $$ where $F$ is an arbitrary functional and where the partition function/path integral is $$\begin{align} Z[J]~:=~& \int \! {\cal D}\phi~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\},\cr S[\phi;J]~:=~&S[\phi]+J_k\phi^k, \end{align}\tag{C}$$ The correspondence (B) follows from the underlying time slicing procedure of path integrals. See e.g. this and this Phys.SE answer.

Now to the main point: Notice how the dictionary (B) naturally talks to $T_{\rm cov}$ rather than $T$: If the functional $F$ doesn't contain time derivatives, it doesn't matter whether we use $T$ or $T_{\rm cov}$. However, if $F$ does contain time derivatives, they get applied outside the correlator, i.e. the time-order is $T_{\rm cov}$.

Then the Schwinger-Dyson (SD) equations becomes $$ \left< \Omega \left| T_{\rm cov}\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~i\hbar\left< \Omega \left| T_{\rm cov}\left\{\frac{\delta F[\phi]}{\delta \phi(x)} \right\}\right| \Omega \right>_J ~, \tag{D}$$ cf. e.g. this Phys.SE post.

In contrast, if we only use the usual time ordering $T$, we do not get the contact term: $$ \left< \Omega \left| T\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~0, \tag{E}$$ because the EOMs are satisfied in quantum average, cf. e.g. this Phys.SE post.

AccidentalFourierTransform's above comment is exactly right: The point is that Srednicki's time-ordering $T$ should be replaced with covariant time-ordering $T_{\rm cov}$, i.e. time-differentiations inside its argument should be taken after/outside the usual time ordering $T$.

This resolves the apparent conflict/contradiction between Srednicki's eqs. (22.23') and (22.24). In other words, the 2-pt function $\left<T\{\phi\phi\} \right>$ can only be the Greens function $\Delta$ if we use $T_{\rm cov}$ instead of $T$ in the SD eq. (22.23).

More generally, the formal correspondence between $$\text{operator formulation} \quad\leftrightarrow\quad \text{path integral formulation}\tag{A} $$ is $$ \left< \Omega \left| T_{\rm cov}\{ F[\phi]\} \right| \Omega \right>_J ~=~ \frac{1}{Z[J]}\int \! {\cal D}\phi~F[\phi]~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\},\tag{B} $$ where the partition function/path integral is $$ Z[J]~:=~ \int \! {\cal D}\phi~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\}.\tag{C}$$ The correspondence (B) follows from the underlying time slicing procedure of path integrals. See e.g. this and this Phys.SE answer.

Then the Schwinger-Dyson (SD) equations becomes $$ \left< \Omega \left| T_{\rm cov}\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(y)}\right\}\right| \Omega \right>_J~=~i\hbar\left< \Omega \left| T_{\rm cov}\left\{\frac{\delta F[\phi]}{\delta \phi(y)} \right\}\right| \Omega \right>_J ~, \tag{D}$$ cf. e.g. this Phys.SE post.

In contrast, if we only use the usual time ordering $T$, we do not get the contact term: $$ \left< \Omega \left| T\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(y)}\right\}\right| \Omega \right>_J~=~0, \tag{E}$$ because the eoms are satisfied in quantum average.

AccidentalFourierTransform's above comment is exactly right: The point is that Srednicki's time-ordering $T$ should be replaced with covariant time-ordering $T_{\rm cov}$, i.e. time-differentiations inside its argument should be taken after/outside the usual time ordering $T$.

This resolves the apparent conflict/contradiction between Srednicki's eqs. (22.23') and (22.24). In other words, the 2-pt function $\left<T\{\phi\phi\} \right>$ can only be the Greens function $\Delta$ if we use $T_{\rm cov}$ instead of $T$ in the SD eq. (22.23).

More generally, the formal correspondence/dictionary between $$\text{operator formulation} \quad\leftrightarrow\quad \text{path integral formulation}\tag{A} $$ is $$\begin{align} \left< \Omega \left| T_{\rm cov}\{ F[\phi]\} \right| \Omega \right>_J ~=~& \frac{1}{Z[J]}\int \! {\cal D}\phi~F[\phi]~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\}\cr ~=~& \frac{1}{Z[J]}F\left[\frac{\hbar}{i}\frac{\delta}{\delta J} \right]Z[J],\end{align}\tag{B} $$ where $F$ is an arbitrary functional and where the partition function/path integral is $$\begin{align} Z[J]~:=~& \int \! {\cal D}\phi~\exp\left\{\frac{i}{\hbar}S[\phi;J]\right\},\cr S[\phi;J]~:=~&S[\phi]+J_k\phi^k, \end{align}\tag{C}$$ The correspondence (B) follows from the underlying time slicing procedure of path integrals. See e.g. this and this Phys.SE answer. Notice how the dictionary (B) naturally talks to $T_{\rm cov}$ rather than $T$.

Then the Schwinger-Dyson (SD) equations becomes $$ \left< \Omega \left| T_{\rm cov}\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~i\hbar\left< \Omega \left| T_{\rm cov}\left\{\frac{\delta F[\phi]}{\delta \phi(x)} \right\}\right| \Omega \right>_J ~, \tag{D}$$ cf. e.g. this Phys.SE post.

In contrast, if we only use the usual time ordering $T$, we do not get the contact term: $$ \left< \Omega \left| T\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~0, \tag{E}$$ because the EOMs are satisfied in quantum average, cf. e.g. this Phys.SE post.