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Rather than going on a definition chase, perhaps the following heuristic derivation of eqs. (F) & (G) is the most convincing/satisfying/instructive. For sufficiently short times $|\Delta t| \ll \tau$, where $\tau$ is some characteristic time scale, i.e. in the diabatic limit, the particle only has time to feel an averaged effect of the potential $V$. So, using methods of Ref. 1, in that limit $|\Delta t| \ll \tau$, the path integral (B) reads

$$\begin{align} K(x_2,t_2;&x_1,t_1)\cr ~=~&\sqrt{\frac{m}{2\pi i\hbar |\Delta t|}} \exp\left\{ \frac{i}{\hbar}\left[ \frac{m}{2} \frac{(\Delta x)^2}{\Delta t}- \langle V\rangle \Delta t \right]\right\}, \end{align}\tag{H} $$

IV) Proof of eq. (G). Note that it is implicitly assumed in eq. (H) that ${\rm Re}(i|\Delta t|)>0$ is slightly positive via the pertinent $i\epsilon$-prescription. Equation (G) then follows directly from eq. (H) via the heat kernel representation

$$ \begin{align} \frac{\partial}{\partial t_2} &K(x_2,t_2;x_1,t_1)\cr ~\stackrel{(H)}{=}~&-\left\{\frac{1}{2\Delta t} +\frac{i}{\hbar}\left[ \frac{m}{2} \left(\frac{\Delta x}{\Delta t}\right)^2+ \langle V\rangle +{\cal O}(\Delta t) \right]\right\} K(x_2,t_2;x_1,t_1),\end{align} \tag{L} $$

$$ \begin{align} \frac{\hbar}{i}\frac{\partial}{\partial x_2} &K(x_2,t_2;x_1,t_1)\cr ~\stackrel{(H)}{=}~&\left\{m \frac{\Delta x}{\Delta t} +{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1), \end{align}\tag{M}$$

$$ \begin{align} \frac{\hbar}{i}\frac{1}{2m}&\frac{\partial^2}{\partial x_2^2} K(x_2,t_2;x_1,t_1) \cr ~\stackrel{(H)}{=}~&\left\{\frac{1}{2\Delta t} +\frac{i}{\hbar} \frac{m}{2} \left(\frac{\Delta x}{\Delta t}\right)^2 +{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1). \end{align} \tag{N}$$

Rather than going on a definition chase, perhaps the following heuristic derivation of eqs. (F) & (G) from formula (H) below is the most convincing/satisfying/instructive. For sufficiently short times $|\Delta t| \ll \tau$, where $\tau$ is some characteristic time scale, i.e. in the diabatic limit, the particle only has time to feel an averaged effect of the potential $V$. So, using methods of Ref. 1, in that limit $|\Delta t| \ll \tau$, the path integral (B) with the $i\epsilon$-prescription reads

$$\begin{align} K(x_2,t_2;&x_1,t_1)\cr ~=~&\sqrt{\frac{m}{2\pi i\hbar (\Delta t-i\epsilon)}} \exp\left\{ \frac{i}{\hbar}\left[ \frac{m}{2} \frac{(\Delta x)^2}{\Delta t-i\epsilon}- \langle V\rangle \Delta t \right]\right\}, \end{align}\tag{H} $$

IV) Proof of eq. (G). Note that it is implicitly assumed in eq. (H) that ${\rm Re}(i\Delta t+\epsilon)>0$ is slightly positive via the pertinent $i\epsilon$-prescription. Equation (G) then follows directly from eq. (H) via the heat kernel representation

$$ \begin{align} \frac{\partial}{\partial t_2} &K(x_2,t_2;x_1,t_1)\cr ~\stackrel{(H)}{=}~&-\left\{\frac{1/2}{\Delta t-i\epsilon} +\frac{i}{\hbar}\left[ \frac{m}{2} \left(\frac{\Delta x}{\Delta t-i\epsilon}\right)^2+ \langle V\rangle +{\cal O}(\Delta t) \right]\right\} K(x_2,t_2;x_1,t_1),\end{align} \tag{L} $$

$$ \begin{align} \frac{\hbar}{i}\frac{\partial}{\partial x_2} &K(x_2,t_2;x_1,t_1)\cr ~\stackrel{(H)}{=}~&\left\{m \frac{\Delta x}{\Delta t-i\epsilon} +{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1), \end{align}\tag{M}$$

$$ \begin{align} \frac{\hbar}{i}\frac{1}{2m}&\frac{\partial^2}{\partial x_2^2} K(x_2,t_2;x_1,t_1) \cr ~\stackrel{(H)}{=}~&\left\{\frac{1/2}{\Delta t-i\epsilon} +\frac{i}{\hbar} \frac{m}{2} \left(\frac{\Delta x}{\Delta t-i\epsilon}\right)^2 +{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1). \end{align} \tag{N}$$

I) Notational issues: Greens function vs. kernel. First of all, be aware that Ref. 1 between eq. (4-27) and eq. (4-28) effectively introduces the retarded Greens function/propagator

rather than the kernel/path integral

Here $\theta$ denotes the Heaviside step function, and the Lagrangian

where we introduced the Schrödinger differential operator

$$ K(x_2,t_2;x_1,t_1) ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0^+. \tag{G}$$ $\Box$

$$\begin{align} K(x_2,t_2;&x_1,t_1)\cr ~=~&\sqrt{\frac{m}{2\pi i\hbar \Delta t}} \exp\left\{ \frac{i}{\hbar}\left[ \frac{m}{2} \frac{(\Delta x)^2}{\Delta t}- \langle V\rangle \Delta t \right]\right\}, \end{align}\tag{H} $$

IV) Proof of eq. (G). Note that it is implicitly assumed in eq. (H) that ${\rm Re}(i\Delta t)>0$ is slightly positive via the pertinent $i\epsilon$-prescription. Equation (G) then follows directly from eq. (H) via the heat kernel representation

of the Dirac delta distribution. $\Box$

I) Notational issues: Greens function vs. kernel. First of all, be aware that Ref. 1 between eq. (4-27) and eq. (4-28) effectively introduces the retarded Greens function/propagator

rather than the kernel/path integral

Here $\theta$ denotes the Heaviside step function, and the Lagrangian

where we introduced the Schrödinger differential operator

$$ K(x_2,t_2;x_1,t_1) ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0. \tag{G}$$ $\Box$

$$\begin{align} K(x_2,t_2;&x_1,t_1)\cr ~=~&\sqrt{\frac{m}{2\pi i\hbar |\Delta t|}} \exp\left\{ \frac{i}{\hbar}\left[ \frac{m}{2} \frac{(\Delta x)^2}{\Delta t}- \langle V\rangle \Delta t \right]\right\}, \end{align}\tag{H} $$

IV) Proof of eq. (G). Note that it is implicitly assumed in eq. (H) that ${\rm Re}(i|\Delta t|)>0$ is slightly positive via the pertinent $i\epsilon$-prescription. Equation (G) then follows directly from eq. (H) via the heat kernel representation

of the Dirac delta distribution. $\Box$

$$\tag{A} G(x_2,t_2;x_1,t_1)~=~\theta(\Delta t)~K(x_2,t_2;x_1,t_1), \qquad \Delta t~:=~t_2-t_1,$$

$$ K(x_2,t_2;x_1,t_1)~=~\langle x_2,t_2 | x_1,t_1 \rangle~=~\langle x_2|U(t_2,t_1)|x_1 \rangle$$ $$ ~=~\int_{x(t_1)=x_1}^{x(t_2)=x_2} \! {\cal D}x~ \exp\left[\frac{i}{\hbar}\int_{t_1}^{t_2} \!dt ~L\right] .\tag{B} $$

$$\tag{C} L~:=~\frac{m}{2}\dot{x}^2-V(x)$$

$$\tag{D} D_2 G(x_2,t_2;x_1,t_1) ~=~\delta(\Delta t)~\delta(\Delta x), \qquad \Delta x~:=~x_2-x_1, $$

$$D_2~:= ~\frac{\partial}{\partial t_2} + \frac{i}{\hbar}\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_2^2}+V(x_2)\right)$$ $$\tag{E} ~=~\frac{\partial}{\partial t_2} + \frac{\hbar}{i}\frac{1}{2m}\frac{\partial^2}{\partial x_2^2}+\frac{i}{\hbar}V(x_2).$$

$$\tag{F} D_2 K(x_2,t_2;x_1,t_1) ~=~0, $$

$$\tag{G} K(x_2,t_2;x_1,t_1) ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0^+. $$ $\Box$

$$\tag{H} K(x_2,t_2;x_1,t_1) ~=~\sqrt{\frac{m}{2\pi i\hbar \Delta t}} \exp\left\{ \frac{i}{\hbar}\left[ \frac{m}{2} \frac{(\Delta x)^2}{\Delta t}- \langle V\rangle \Delta t \right]\right\}, $$

$$\tag{I} \langle V\rangle ~=~ V\left(\frac{x_1+x_2}{2}\right)+{\cal O}(\Delta x) ~=~ V(x_2)+{\cal O}(\Delta x)~=~ V(x_1)+{\cal O}(\Delta x). $$

$$\tag{J} \delta(x)~=~ \lim_{|\alpha|\to \infty} \sqrt{\frac{\alpha}{\pi}} e^{-\alpha x^2}, \qquad {\rm Re}(\alpha)~>~0, $$

Lemma. For sufficiently small times $|\Delta t| \ll \tau$, the path integral (H) satisfies $$\tag{K} D_2 K(x_2,t_2;x_1,t_1) ~=~{\cal O}(\Delta t). $$

$$ \frac{\partial}{\partial t_2} K(x_2,t_2;x_1,t_1)~\stackrel{(H)}{=}~-\left\{\frac{1}{2\Delta t} +\frac{i}{\hbar}\left[ \frac{m}{2} \left(\frac{\Delta x}{\Delta t}\right)^2+ \langle V\rangle +{\cal O}(\Delta t) \right]\right\} K(x_2,t_2;x_1,t_1),\tag{L} $$

$$ \frac{\hbar}{i}\frac{\partial}{\partial x_2} K(x_2,t_2;x_1,t_1) ~\stackrel{(H)}{=}~\left\{m \frac{\Delta x}{\Delta t} +{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1), \tag{M}$$

$$ \frac{\hbar}{i}\frac{1}{2m}\frac{\partial^2}{\partial x_2^2} K((x_2,t_2;x_1,t_1) ~\stackrel{(H)}{=}~\left\{\frac{1}{2\Delta t} +\frac{i}{\hbar} \frac{m}{2} \left(\frac{\Delta x}{\Delta t}\right)^2 +{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1). \tag{N}$$

$$ \tag{O} \left\{V(x_2)-\langle V\rangle \right\}K(x_2,t_2;x_1,t_1) ~\stackrel{(I)}{=}~{\cal O}(\Delta x) K(x_2,t_2;x_1,t_1) ~\stackrel{(G)}{=}~{\cal O}(\Delta t), $$

$$\tag{2-31} K(x_2,t_2;x_1,t_1)~=~\int_{\mathbb{R}} \! dx_3~ K(x_2,t_2;x_3,t_3)~K(x_3,t_3;x_1,t_1), $$

$$ D_2 K(x_2,t_2;x_1,t_1)~\stackrel{(2-31)}{=}~\int_{\mathbb{R}} \! dx_3~ D_2 K(x_2,t_2;x_3,t_3)~K(x_3,t_3;x_1,t_1)$$ $$ ~\stackrel{(K)}{=}~\int_{\mathbb{R}} \! dx_3~ {\cal O}(t_2-t_3)~K(x_3,t_3;x_1,t_1)~=~{\cal O}(t_2-t_3) .\tag{P}$$

$$\begin{align} G(x_2,t_2;x_1,t_1)~=~&\theta(\Delta t)~K(x_2,t_2;x_1,t_1), \cr \Delta t~:=~&t_2-t_1,\end{align}\tag{A}$$

$$\begin{align} K(x_2,t_2;&x_1,t_1)\cr ~=~&\langle x_2,t_2 | x_1,t_1 \rangle\cr ~=~&\langle x_2|U(t_2,t_1)|x_1 \rangle\cr ~=~&\int_{x(t_1)=x_1}^{x(t_2)=x_2} \! {\cal D}x~ \exp\left[\frac{i}{\hbar}\int_{t_1}^{t_2} \!dt ~L\right] .\end{align}\tag{B} $$

$$ L~:=~\frac{m}{2}\dot{x}^2-V(x)\tag{C}$$

$$\begin{align} D_2 G(x_2,t_2;x_1,t_1) ~=~&\delta(\Delta t)~\delta(\Delta x), \cr \Delta x~:=~&x_2-x_1, \end{align}\tag{D} $$

$$\begin{align}D_2~:= ~&\frac{\partial}{\partial t_2} + \frac{i}{\hbar}\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_2^2}+V(x_2)\right)\cr ~=~&\frac{\partial}{\partial t_2} + \frac{\hbar}{i}\frac{1}{2m}\frac{\partial^2}{\partial x_2^2}+\frac{i}{\hbar}V(x_2).\end{align}\tag{E}$$

$$ D_2 K(x_2,t_2;x_1,t_1) ~=~0, \tag{F} $$

$$ K(x_2,t_2;x_1,t_1) ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0^+. \tag{G}$$ $\Box$

$$\begin{align} K(x_2,t_2;&x_1,t_1)\cr ~=~&\sqrt{\frac{m}{2\pi i\hbar \Delta t}} \exp\left\{ \frac{i}{\hbar}\left[ \frac{m}{2} \frac{(\Delta x)^2}{\Delta t}- \langle V\rangle \Delta t \right]\right\}, \end{align}\tag{H} $$

$$\begin{align} \langle V\rangle ~=~& V\left(\frac{x_1+x_2}{2}\right)+{\cal O}(\Delta x)\cr ~=~& V(x_2)+{\cal O}(\Delta x)\cr ~=~& V(x_1)+{\cal O}(\Delta x). \end{align}\tag{I}$$

$$ \delta(x)~=~ \lim_{|\alpha|\to \infty} \sqrt{\frac{\alpha}{\pi}} e^{-\alpha x^2},\qquad {\rm Re}(\alpha)~>~0, \tag{J} $$

Lemma. For sufficiently small times $|\Delta t| \ll \tau$, the path integral (H) satisfies $$ D_2 K(x_2,t_2;x_1,t_1) ~=~{\cal O}(\Delta t). \tag{K}$$

$$ \begin{align} \frac{\partial}{\partial t_2} &K(x_2,t_2;x_1,t_1)\cr ~\stackrel{(H)}{=}~&-\left\{\frac{1}{2\Delta t} +\frac{i}{\hbar}\left[ \frac{m}{2} \left(\frac{\Delta x}{\Delta t}\right)^2+ \langle V\rangle +{\cal O}(\Delta t) \right]\right\} K(x_2,t_2;x_1,t_1),\end{align} \tag{L} $$

$$ \begin{align} \frac{\hbar}{i}\frac{\partial}{\partial x_2} &K(x_2,t_2;x_1,t_1)\cr ~\stackrel{(H)}{=}~&\left\{m \frac{\Delta x}{\Delta t} +{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1), \end{align}\tag{M}$$

$$ \begin{align} \frac{\hbar}{i}\frac{1}{2m}&\frac{\partial^2}{\partial x_2^2} K(x_2,t_2;x_1,t_1) \cr ~\stackrel{(H)}{=}~&\left\{\frac{1}{2\Delta t} +\frac{i}{\hbar} \frac{m}{2} \left(\frac{\Delta x}{\Delta t}\right)^2 +{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1). \end{align} \tag{N}$$

$$ \begin{align} \left\{V(x_2)-\langle V\rangle \right\}K(x_2,t_2;x_1,t_1) ~\stackrel{(I)}{=}~&{\cal O}(\Delta x) K(x_2,t_2;x_1,t_1)\cr ~\stackrel{(G)}{=}~&{\cal O}(\Delta t),\end{align} \tag{O}$$

$$ \begin{align} K(x_2,t_2;&x_1,t_1)\cr ~=~&\int_{\mathbb{R}} \! dx_3~ K(x_2,t_2;x_3,t_3)~K(x_3,t_3;x_1,t_1),\end{align} \tag{2-31}$$

$$ \begin{align}D_2 K(x_2,t_2;&x_1,t_1)\cr~\stackrel{(2-31)}{=}~&\int_{\mathbb{R}} \! dx_3~ D_2 K(x_2,t_2;x_3,t_3)~K(x_3,t_3;x_1,t_1) \cr ~\stackrel{(K)}{=}~&\int_{\mathbb{R}} \! dx_3~ {\cal O}(t_2-t_3)~K(x_3,t_3;x_1,t_1)\cr ~=~&{\cal O}(t_2-t_3) .\end{align}\tag{P}$$