$$ \newcommand{\ddt}[1]{\frac{d #1}{dt}} $$ This is quoted from Feynman's lectures' The Schrödinger equation:
In Chapter 8, we described how states varied in time in terms of the Hamiltonian $H_{ij}.$ We saw that the time variation of the various amplitudes was given in terms of the matrix equation $$iℏ\ddt{C_i}=\sum_jH_{ij}C_j.\tag{16.49}$$ This equation says that the time variation of each amplitude $C_i$ is proportional to all of the other amplitudes $C_j,$ with the coefficients $H_{ij}.$ How would we expect Eq. $(16.49)$ to look when we are using the continuum of base states $|x⟩?$ Let’s first remember that Eq. $(16.49)$ can also be written as $$iℏ\ddt{⟨i|ψ⟩}=\sum_j ⟨i|\hat{H}|j⟩⟨j|ψ⟩.$$ Now it is clear what we should do. For the $x$-representation we would expect $$iℏ\frac{∂}{∂t}⟨x|ψ⟩=∫⟨x|\hat{H}|x'⟩⟨x'|ψ⟩dx'.\tag{16.50}$$ The sum over the base states $|j⟩,$ gets replaced by an integral over $x'.$ Since $⟨x|\hat{H}|x'⟩$ should be some function of $x$ and $x',$ we can write it as $H(x,x')$—which corresponds to $H_{ij}$ in Eq. $(16.49).$ Then Eq. $(16.50)$ is the same as $$iℏ\frac{∂}{∂t}ψ(x)=∫H(x,x')ψ(x')dx'\tag{16.51}$$ with $$H(x,x')≡⟨x|\hat{H}|x'⟩.$$ According to Eq. $(16.51),$ the rate of change of $ψ$ at $x$ would depend on the value of $ψ$ at all other points $x';$ the factor $H(x,x')$ is the amplitude per unit time that the electron will jump from $x'$ to $x.$ It turns out in nature, however, that this amplitude is zero except for points $x'$ very close to $x.$ This means—as we saw in the example of the chain of atoms at the beginning of the chapter, Eq. $(16.12)$—that the right-hand side of Eq. $(16.51)$ can be expressed completely in terms of $ψ$ and the derivatives of $ψ$ with respect to $x,$ all evaluated at the position $x.$ For a particle moving freely in space with no forces, no disturbances, the correct law of physics is $$∫H(x,x')ψ(x')dx'=−\frac{ℏ^2}{2m}\frac{∂^2}{∂x^2}ψ(x).$$ Where did we get that from? Nowhere. It’s not possible to derive it from anything you know. It came out of the mind of Schrödinger, invented in his struggle to find an understanding of the experimental observations of the real world. You can perhaps get some clue of why it should be that way by thinking of our derivation of Eq. $(16.12)$ which came from looking at the propagation of an electron in a crystal. Of course, free particles are not very exciting. What happens if we put forces on the particle? Well, if the force of a particle can be described in terms of a scalar potential $V(x)$—which means we are thinking of electric forces but not magnetic forces—and if we stick to low energies so that we can ignore complexities which come from relativistic motions, then the Hamiltonian which fits the real world gives $$∫H(x,x')ψ(x')dx'=−\frac{ℏ^2}{2m}\frac{∂^2}{∂x^2}ψ(x)+V(x)ψ(x).\tag{16.52}$$ Again, you can get some clue as to the origin of this equation if you go back to the motion of an electron in a crystal, and see how the equations would have to be modified if the energy of the electron varied slowly from one atomic site to the other—as it might do if there were an electric field across the crystal. Then the term $E_0$ in Eq. $(16.7)$ would vary slowly with position and would correspond to the new term we have added in $(16.52).$ [You may be wondering why we went straight from Eq. $(16.51)$ to Eq. $(16.52)$ instead of just giving you the correct function for the amplitude $H(x,x')=⟨x|\hat{H}|x'⟩.$ We did that because $H(x,x')$ can only be written in terms of strange algebraic functions, although the whole integral on the right-hand side of Eq. $(16.51)$ comes out in terms of things you are used to. If you are really curious, $H(x,x')$ can be written in the following way: $$H(x,x')=−\frac{ℏ^2}{2m}δ''(x−x')+V(x)δ(x−x'),$$ where $δ''$ means the second derivative of the delta function. This rather strange function can be replaced by a somewhat more convenient algebraic differential operator, which is completely equivalent: $$H(x,x')=\left\{−\frac{ℏ^2}{2m}\frac{∂^2}{∂x^2}+V(x)\right\}δ(x−x').$$ We will not be using these forms, but will work directly with the form in Eq. $(16.52).$]
Feynman, from nowhere, jot down $H(x,x')=−\dfrac{ℏ^2}{2m}δ''(x−x')+V(x)δ(x−x')$; no deduction, no proof:( Can anyone please explain how Feynman wrote this? Meant to say, what is the proof?
Also, I knew from the beginning that $H(x,x')$ is the amplitude per unit time that the electron will jump from $x'$ to $x.$ But why did Feynman say 'It turns out in nature, however, that this amplitude is zero except for points $x'$ very close to $x.$'?
Your last equation is missing the parentheses, $$ H(x,x') = \left\{ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right\}\delta(x-x'). $$ We can see that this is equivalent to the other form of $H(x,x')$, since $$ \frac{\partial^2}{\partial x^2}\delta(x-x') = \delta''(x-x'). $$ As for where this comes from, Feynman said it himself in that excerpt:
This is simply the form $H(x,x')$ takes in order for (16.52) to be satisfied:
\begin{align*} \int H(x,x')\psi(x')dx' &= \int\left\{ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right\}\delta(x-x')\psi(x')dx' \\ &= \left\{ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right\} \int \delta(x-x')\psi(x')dx' \\ &= \left\{ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right\}\psi(x) \\ &= -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x) + V(x)\psi(x). \end{align*}
It is required to make the theory match the observations.
As for why Feynman said that $H(x,x')$ is zero except for $x,x'$ very close to each other; again, it is because this matches our observations of what happens in experiments. Physically, it means that it is unlikely/impossible for electrons to make large jumps.