Check out my answer to a related question here. Notice that the next equation Sakurai gives makes the key difference.
$$
K({\mathbf x}^{\prime\prime},t; {\mathbf x}^\prime, t_0) = 0 \quad\text{ for } t<t_0\,.
$$
This is implicitly the $\Theta(t - t_0)$ function imposing time ordering that I mention. It makes the difference between the Dirac $\delta$ driving terms and not. It also explains the coefficient you ask about.
$$
-i\hbar\,\frac{d}{dt}\,\Theta(t - t_0) = -i\hbar\,\delta(t - t_0)
$$
The $\delta$-function on the spatial points comes from the answer I linked to.
First, the term "propagator" is usually defined as the Green's function of the first type, not the second type, i.e. as a solution to the diffential equation $\hat L G = \delta$.
At any rate, those definitions are ultimately equivalent – when the details are correctly written down – because the Green's function defined as the correlator in the second definition obeys the first differential equation.
The differential operator $\hat L$ is what appears in the linearized equations of motion for the field, in this case $\psi(x_1,t_1)$, and it only acts on $\psi(x_1,t_1)$, not $\psi^\dagger (x_2,t_2)$.
The time-ordering operator $T$ may be written in terms of the step function
$$ T(\psi \psi^\dagger) = \psi \psi^\dagger\cdot \theta(t_1-t_2) - \psi^\dagger \psi\cdot\theta(-t_1+t_2)$$
where $\psi$ is always at $x_1,t_1$ and $\psi^\dagger$ is at $x_2,t_2$. Now, ask what happens when you act with $\hat L$ on the right hand side of the displayed equation above.
By the Leibniz rule, there are the terms with $\hat L \psi = 0$. It vanishes by the equations of motion. But there are extra terms where $\hat L$ acts on the step functions.
The operator $\hat L$ contains the term that differentiates with respect to $t_1$ multiplied by a coefficient $C$. This turns $\theta(t_1-t_2)$ to $\delta(t_1-t_2)$. The same occurs in the next term, but with the opposite sign which cancels the sign that was already there. So the extra terms are
$$ \hat L T(\psi \psi^\dagger) = C\delta(t_1-t_2) (\psi \psi^\dagger + \psi^\dagger \psi)$$
I got two terms because there were two terms. However, these two terms exactly combine to the anticommutator of $\psi$ and $\psi^\dagger$ which only needs to be evaluated for $t_1=t_2$, the equal-time anticommutator, and the result is $D\cdot \delta(x_1-x_2)$.
That's why the action of $\hat L$ on the correlator ends up being $CD\delta(t_1-t_2)\delta(x_1-x_2)$ where the constants $C,D$ are mostly just factors of $i$ etc.
For bosonic fields, $\hat L$ has the second derivative with respect to time. One of the derivatives has the fate as above, the other one turns the other $\phi$, which plays the role of $\psi^\dagger$, into $\partial_t \phi$ which is the canonical momentum, and it's the right variable that has the $\delta$-function-like commutator. Also, the intermediate sign is the opposite one but the result is the same, some $CD\cdot\delta\cdot\delta$.
Best Answer
Your question has been answered again and again, and again, albeit indirectly and elliptically--I'll just be more direct and specific. The point is you skipped variables: in this case, t, and so the expression you wrote ("according to books, $\hat{L}K(x;x') = \delta(x-x')~$"), is nonsense, as you already properly found out; unless you included t in the generalized coordinates, but then again the convolution preceding it is not right.
It is all a deplorable misunderstanding, caused by sloppy language in the community. The WP article gets it right. The retarded Green's function G is the inverse of $\hat{L}=i\hbar\partial_t-H$, $$ \hat{L}G(x,t;x',t') = \delta(t-t')\delta(x-x')~, $$ and it is not exactly the propagator. (Together with its advanced twin, they comprise a hyperformal unitary time evolution operator, of no practical concern here.)
Consider $$G\equiv \frac{1}{i\hbar} \theta(t-t') K(x,t;x',t'), $$ and posit $K(x,t;x',t)=\delta(x-x')$. In that case, the propagator K turns out to be the kernel (null eigenfunction) of $\hat{L}$, simply because the time derivative in $\hat{L}$ acting on the step function $\theta(t-t')$, yields a $\delta (t-t')$ and hence a $\delta(x-x')$ when acting on K by the above posit. These then cancel the two deltas on the r.h.side, and leave only $$ \theta(t-t'\!) ~~ \hat{L}K(x,t;x',t') = 0 $$ behind.
So K is the fundamental solution of the homogeneous equation, the real TDSE (recall you never wish to solve the inhomogeneous TDSE!); and all works out.
Without loss of generality, take $t'=0$, so write $K(x,t;x')$. Consequently $$ \psi(x,t)=\int dx' K(x,t;x') u(x') $$ is a null eigenfunction of $\hat{L}$ with initial condition $\psi(x,0)=u(x)$, which, in turn, justifies the posit.
That is, the above integral is the most general superposition of the fundamental solutions corresponding to all possible I.C.s
Illustrating the above with the free particle, $$ \left(i\hbar\partial_t + \frac{\hbar^2 \partial^2_x}{2m}\right) K(x,t;x')= 0 $$ yields $$ K(x,t;x')=\frac{1}{2\pi}\int_{-\infty}^{+\infty}dk\,e^{ik(x-x')} e^{-\frac{i\hbar k^2 t}{2m}}=\left(\frac{m}{2\pi i\hbar t}\right)^{\frac{1}{2}}e^{-\frac{m(x-x')^2}{2i\hbar t}}, $$ which satisfies the I.C. posit.
You appear to have verified the solution of the above homogeneous equation already, $$ \left ( i\hbar\left( -\frac{1}{2t} +\frac{m(x-x')^2}{2i\hbar t^2}\right) +\frac{\hbar^2}{2m}\left ( -\frac{m}{i\hbar t } - \frac{m^2(x-x')^2}{\hbar^2 t^2} \right) \right ) K =0, $$ alright.
It is also straightforward to likewise find the propagator for the oscillator (quadratic potential), so then solve for the magnificent Mehler kernel (1866), $$ K(x,t;x')=\left(\frac{m\omega}{2\pi i\hbar \sin \omega t}\right)^{\frac{1}{2}}\exp\left(-\frac{m\omega((x^2+x'^2)\cos\omega t-2xx')}{2i\hbar \sin\omega t}\right) ~,$$ which, analytically continued, precedes this QM problem by more than half a century...