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...
The Green's function satisfies
$$
\nabla_{\bf x}^2 G({\bf x}) = \delta ({\bf x}) \tag{1}
$$
where ${\bf x}=(x,y,z)$ is a 3-vector. The general solution to this equation that vanishes at infinity is given by the Fourier transform
$$
G({\bf x}) = \int d{\bf k} \frac{1}{|{\bf k}|^2} e^{ i {\bf k} \cdot {\bf x} }
$$
It follows from this that $G({\bf x})$ is rotationally symmetric (prove this yourself!) so we must have $G({\bf x}) = G(r)$ where $r=|{\bf x}|$.
Now, to determine $G(r)$, we can either just calculate the integral above or we can do what the author did. We integrate (1) inside a ball of radius $R$. This $R$ does not mean anything. It is a mathematical tool employed to find out what the function $G(r)$ is. We find
$$
\int_B dx dy dz \nabla_{\bf x}^2 G({\bf x}) = \int_B dx dy dz \delta ({\bf x}) = 1
$$
We now move to spherical coordinates
$$
1 = \int_0^R dr r^2 \int_0^\pi d\theta \sin \theta \int_0^{2\pi} d \phi \frac{1}{r^2} \frac{d}{dr} ( r^2 G'(r) )
$$
All the integrals are super easy to evaluate and we find
$$
1 = 4\pi ( r^2 G'(r) ) |_{r=R} = 4\pi R^2 G'(R) .
$$
Now, this is true for any value of the the radius $R$ so we can setup another differential equation
$$
4\pi R^2 G'(R) = 1 \implies G(R) = - \frac{1}{4\pi R} + C
$$
The constant can be fixed by requiring that $G(R) \to 0$ as $R \to \infty$ so we have $C=0$. In summary,
$$
G({\bf x}) = G(r) = - \frac{1}{4\pi r} = - \frac{1}{4\pi |{\bf x} | } .
$$
Best Answer
You can see from the expression $$RG_k(R)=Ae^{ikR} + Be^{-ikR},$$ that the solution depends only on the combination $kR$. You know additionally that the Dirac delta distribution will only have an impact when $R$ goes to zero as stated. When evaluating that limit the value of $k$ becomes irrelevant (as long as it is finite) since $kR\rightarrow 0$ anyways. In particular you can make your life easier by using a specific value of $k$, namely 0, for the original differential equation. This "trick" is what leads to the Poisson equation and Jackson describes very briefly.
Observe this is simply allowing you to find the normalization by comparison with the known case from electrostatics.