understanding of the classical definition of Green's function I learn the classical definition of Green's function from Hunter's Applied Analysis.
Consider the second-order ordinary differential operators $A$ of the form 
$$Au=au''+bu'+cu,$$
where $a,b$ and $c$ are sufficiently smooth functions on $[0,1]$.  
Think about the Dirichlet boundary value problem for the second-order differential operator $A$ defined above:
$$Au=f,\qquad u(0)=u(1)=0,\qquad \tag{10.9}$$
where $f:[0,1]\to{\mathbb C}$ is a given continuous function.
The author gives a heuristic discussion using Dirac delta function:
the Green's function $g(x,y)$ associated with the boundary value problem in (10.9) is the solution of the following problem:
$$Ag(x,y)=\delta(x-y),\qquad g(0,y)=g(1,y)=0.\qquad \tag{10.13}$$
He reformulates $(10.13)$ in classical, pointwise terms. The book said that we want $g(x,y)$ to satisfy the homogeneous ODE(as a function of $x$) when $x\neq y$, and we want the jump in $a(x)g_x(x,y)$ across $x=y$ to equal one in order to obtain a delta function after taking a second $x$-derivative. We therefore make the following definition:† 


† A function $g:[0,1]\times[0,1]\to{\mathbb C}$ is a Green's function for (10.9) if it satisfies the following conditions.
(a) The function $g(x,y)$ is continuous on the square $0\leq x,y\leq 1$, and twice continuously differentiable with respect to $x$ on the triangles $0\leq x\leq y\leq 1$ and $0\leq y\leq x\leq 1$, meaning that the partial derivatives exist in the interiors of the triangles and extend to continuous functions on the closures. The left and right limits of the partial derivatives on $x=y$ are not equal, however. 
(b) The function $g(x,y)$ satisfies the ODE with respect to $x$ and the boundary conditions:
  $$\begin{align}
Ag=0\qquad \text{in}~0<x<y<1~\text{and}~0<y<x<1,\\
g(0,y)=g(1,y)=0\qquad\text{for}~0\leq y\leq 1.
\end{align}
$$
  (c) The jump in $g_x$ across the line $x=y$ is given by
  $$g_x(y^+,y)-g_x(y^-,y)=\frac{1}{a(y)}$$
  where the subscript $x$ denotes a partial derivative with respect to the first variable in $g(x,y)$, and 
  $$g_x(y^+,y)=\lim_{x\to y^+}g_x(x,y),\qquad g_x(y^-,y)=\lim_{x\to y^-}g_x(x,y).$$


The words in bold---
...we want the jump in $a(x)g_x(x,y)$ across $x=y$ to equal one in order to obtain a delta function after taking a second $x$-derivative
refer to condition (c) in the definition above. 
Here is my question:
How can one get 
$Ag(x,y)=\delta(x-y)$
from
$$g_x(y^+,y)-g_x(y^-,y)=\frac{1}{a(y)}?$$
Added:
The confusion is that I don't know what the words in bold mean. Finally, we want $$Ag(x,y)=\delta(x-y),$$ but what's the relation between "taking a second $x$-derivative" of $a(x)g_x(x,y)$ and $Ag(x,y)$?
 A: Let $f:(a,b]\to\mathbb{R}$ and $g:[b,c)\to\mathbb{R}$ be $C^1$-functions. The derivative of $F:(a,c)\to\mathbb{R}$ defined by
$$F(x):=\begin{cases}f(x) & x<b \cr g(x) & x>b\end{cases}$$
equals $\tilde F + (g(b)-f(b))\delta_b$ where $\tilde F$ is the classical derivative
$$\tilde F(x):=\begin{cases}f'(x) & x<b \cr g'(x) & x>b\end{cases}$$
in the distributional sense. In other words, if a function has a jump, then its derivative is a delta distribution which measures the height of the jump (in addition to the classical derivative).
See also my answer to this question to get a very quick overview on distributions, and what it means for a function to have a distribution or a measure as its derivative (note that the delta distribution is actually a measure).
Edit: Here is a heuristic reason of why this is true. The function $F$ is defined piecewise by $f$ on the left side of $b$ and $g$ on the right side of $b$ (the value at $b$ precisely is not important, you can think of the function as a graph which has a vertical line at $b$), but the two pieces don't fit together at $b$. What could the derivative of $F$ reasonably be? If you just ignore the discontinuity at $b$ and differentiate $f$ and $g$ seperately, you get $\tilde F$, but you see that it's not a good candidate for a derivative as soon as you test the fundamental theorem of calculus: Let $x\in (a,b)$ and $y\in (b,c)$, then
$$\int_x ^y \tilde F(t) dt=\int_x ^b f'(t) dt + \int_b ^y g'(t) dt=f(b)-f(x)+g(y)-g(b)$$
where it should be $g(y)-f(x)$. So we are off by $g(b)-f(b)$ which is the height of the jump, so we correct it by adding this number times $\delta_b$ to it. $\delta_b$ has the property that it's zero outside of $b$, and
$$\int_x ^y \delta_b(t)=1$$
Of course such a function does not exist, that's why it's heuristic. Making this rigorous requires the theory of distributions.
A: Recall that, by definition, we require continuity of the Green's function at $x=y$. 
Consider the ODE involving our Green's function,
$$A[g] = a_2(x)\frac{\mathrm d^2 g(x,y)}{\mathrm{dx}^2}+a_1(x)\frac{\mathrm{d}g(x,y)}{\mathrm{dx}} + a_0(x)g(x,y) = \mathrm{\delta}(x-y)\ .$$
We now integrate the equation around the point of continuity $x=y$ of $g$ by taking the following limit
$$\lim_{\varepsilon \to 0} \int_{x = y-\varepsilon}^{y+\varepsilon}A[g]\ dx = \lim_{\varepsilon \to 0} \int_{y-\varepsilon}^{y+\varepsilon}\mathrm{\delta}(x-y)\ dx \ .$$
The right hand side evaluates to $1$. 
On the left hand side, everything vanishes except the second derivative term. (You can check this by using integration by parts and exploiting the fact that $g$ is continuous).
That being said, using integration by parts, we can write
$$ \lim_{\varepsilon \to 0}\ a_2(x)\frac{\mathrm{d}g(x,y)}{\mathrm{dx}} \Bigg|_{x=y-\epsilon}^{y+\varepsilon}-\int_{y-\varepsilon}^{y+\varepsilon}\frac{\mathrm{d}a_2(x)}{\mathrm{dx}}\frac{\mathrm{d}g(x,y)}{\mathrm{dx}}dx = 1\ .$$
Again, by using integration by parts, you can show that the second term vanishes because $g$ is continuous in $x=y$. 
Hence, we now arrive at the "jump condition",
$$\lim_{\varepsilon \to 0}\ a_2(x)\frac{\mathrm{d}g(x,y)}{\mathrm{dx}} \Bigg|_{x=y-\epsilon}^{y+\varepsilon}  = 1\ .$$
Upon taking the limit, we obtain your expression 
$$g_x(y^+,y)-g_x(y^{-},y) = \frac{1}{a_2(x)}\ . $$
