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
$$\begin{align} G(x_2,t_2;x_1,t_1)~=~&\theta(\Delta t)~K(x_2,t_2;x_1,t_1), \cr
\Delta t~:=~&t_2-t_1,\end{align}\tag{A}$$
rather than the kernel/path integral
$$\begin{align} K(x_2,t_2;&x_1,t_1)\cr
~=~&\langle x_2,t_2 | x_1,t_1 \rangle\cr
~=~&\langle x_2|U(t_2,t_1)|x_1 \rangle\cr ~=~&\int_{x(t_1)=x_1}^{x(t_2)=x_2} \! {\cal D}x~ \exp\left[\frac{i}{\hbar}\int_{t_1}^{t_2} \!dt ~L\right] .\end{align}\tag{B} $$
Here $\theta$ denotes the Heaviside step function, and the Lagrangian
$$ L~:=~\frac{m}{2}\dot{x}^2-V(x)\tag{C}$$
is the Lagrangian for a non-relativistic point particle in 1 dimension with a potential $V$.
However, Ref. 1 confusingly denotes the Greens function $G$ with the same letter $K$ as the kernel! See also e.g. this and this Phys.SE posts. Therefore the eq. (4-29) in Ref. 1, which OP asks about, is better written as
$$\begin{align} D_2 G(x_2,t_2;x_1,t_1) ~=~&\delta(\Delta t)~\delta(\Delta x), \cr \Delta x~:=~&x_2-x_1, \end{align}\tag{D} $$
where we introduced the Schrödinger differential operator
$$\begin{align}D_2~:= ~&\frac{\partial}{\partial t_2} + \frac{i}{\hbar}\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_2^2}+V(x_2)\right)\cr
~=~&\frac{\partial}{\partial t_2} + \frac{\hbar}{i}\frac{1}{2m}\frac{\partial^2}{\partial x_2^2}+\frac{i}{\hbar}V(x_2).\end{align}\tag{E}$$
II) Proof of eq. (D). The sought-for eq. (D) follows directly from eq. (A) together with the following two properties (F) & (G) of the kernel $K$:
$$ D_2 K(x_2,t_2;x_1,t_1) ~=~0, \tag{F} $$
and
$$ K(x_2,t_2;x_1,t_1) ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0. \tag{G}$$
$\Box$
III) So we can reformulate OP's question as follows.
Why the path integral (B) satisfies eqs. (F) & (G)?
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
$$\begin{align} K(x_2,t_2;&x_1,t_1)\cr
~=~&\sqrt{\frac{m}{2\pi i\hbar (\Delta t-i\epsilon)}}
\exp\left\{ \frac{i}{\hbar}\left[
\frac{m}{2} \frac{(\Delta x)^2}{\Delta t-i\epsilon}- \langle V\rangle \Delta t
\right]\right\}, \end{align}\tag{H} $$
where the averaged potential is of the form
$$\begin{align} \langle V\rangle ~=~& V\left(\frac{x_1+x_2}{2}\right)+{\cal O}(\Delta x)\cr
~=~& V(x_2)+{\cal O}(\Delta x)\cr
~=~& V(x_1)+{\cal O}(\Delta x). \end{align}\tag{I}$$
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
$$ \delta(x)~=~ \lim_{|\alpha|\to \infty} \sqrt{\frac{\alpha}{\pi}} e^{-\alpha x^2},\qquad
{\rm Re}(\alpha)~>~0, \tag{J} $$
of the Dirac delta distribution. $\Box$
V) Proof of eq. (F) for sufficiently small times $|\Delta t| \ll \tau$. It is a straightforward to check that eq. (H) satisfies the eq. (F) modulo contributions that vanish as $\Delta t\to 0$, cf. the following Lemma. $\Box$
Lemma. For sufficiently small times $|\Delta t| \ll \tau$, the path integral (H) satisfies
$$ D_2 K(x_2,t_2;x_1,t_1) ~=~{\cal O}(\Delta t). \tag{K}$$
Sketched proof of eq. (K): Straightforward differentiation yields
$$ \begin{align} \frac{\partial}{\partial t_2} &K(x_2,t_2;x_1,t_1)\cr
~\stackrel{(H)}{=}~&-\left\{\frac{1/2}{\Delta t-i\epsilon} +\frac{i}{\hbar}\left[
\frac{m}{2} \left(\frac{\Delta x}{\Delta t-i\epsilon}\right)^2+ \langle V\rangle
+{\cal O}(\Delta t) \right]\right\} K(x_2,t_2;x_1,t_1),\end{align} \tag{L} $$
$$ \begin{align} \frac{\hbar}{i}\frac{\partial}{\partial x_2} &K(x_2,t_2;x_1,t_1)\cr
~\stackrel{(H)}{=}~&\left\{m \frac{\Delta x}{\Delta t-i\epsilon}
+{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1), \end{align}\tag{M}$$
$$ \begin{align} \frac{\hbar}{i}\frac{1}{2m}&\frac{\partial^2}{\partial x_2^2} K(x_2,t_2;x_1,t_1) \cr
~\stackrel{(H)}{=}~&\left\{\frac{1/2}{\Delta t-i\epsilon} +\frac{i}{\hbar}
\frac{m}{2} \left(\frac{\Delta x}{\Delta t-i\epsilon}\right)^2
+{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1). \end{align} \tag{N}$$
Also note that
$$ \begin{align} \left\{V(x_2)-\langle V\rangle \right\}K(x_2,t_2;x_1,t_1)
~\stackrel{(I)}{=}~&{\cal O}(\Delta x) K(x_2,t_2;x_1,t_1)\cr
~\stackrel{(G)}{=}~&{\cal O}(\Delta t),\end{align} \tag{O}$$
due to eqs. (I) and (G). The Lemma now follows by combining eqs. (E), (L), (N) & (O). $\Box$
VI) Proof of eq. (F) for large $\Delta t$. We use the path integral property
$$ \begin{align}
K(x_2,t_2;&x_1,t_1)\cr
~=~&\int_{\mathbb{R}} \! dx_3~ K(x_2,t_2;x_3,t_3)~K(x_3,t_3;x_1,t_1),\end{align} \tag{2-31}$$
which is independent of the instant $t_3$. We now pinch the instant $t_3$ sufficiently close to the instant $t_2$, so that we can approximate the path integral $K(x_2,t_2;x_3,t_3)$ by the analog of eq. (H). If we apply the operator $D_2$ on the kernel, we get
$$ \begin{align}D_2 K(x_2,t_2;&x_1,t_1)\cr~\stackrel{(2-31)}{=}~&\int_{\mathbb{R}} \! dx_3~ D_2 K(x_2,t_2;x_3,t_3)~K(x_3,t_3;x_1,t_1) \cr ~\stackrel{(K)}{=}~&\int_{\mathbb{R}} \! dx_3~ {\cal O}(t_2-t_3)~K(x_3,t_3;x_1,t_1)\cr
~=~&{\cal O}(t_2-t_3) .\end{align}\tag{P}$$
Since the lhs. of eq. (P) does not depend on $t_3$, we conclude that it is zero. Hence eq. (F) also holds for large $\Delta t$ as well. $\Box$
References:
- R.P. Feynman & A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965.
