How this series is produced? 
All derivatives can be expressed exactly in terms of infinite series
  of forward, backward, or central-differences. For example
  $$\frac{\partial^2 U}{\partial x^2}=\frac{1}{h^2} ( \delta_x^2 U - \frac{1}{12} \delta_x^4 U + \frac{1}{90} \delta_x^6 U + \cdots ) \tag{$*$}$$ where the subscript $x$ denotes differencing in the
  $x$-direction and the central-differences are defined by $$\delta_x U_{i,\,j} = U_{i+\frac{1}{2},\,j}-U_{i-\frac{1}{2},\,j}$$ and 
  $$\delta_x^2 U_{i,\,j} = \delta_x(\delta_x U_{i,\,j})=U_{i+1,\,j}-2U_{i,\,j}+U_{i-1,\,j}, \quad \text{etc.}$$


In the quote above, which is from a book, it has been said that differential equations can be written as infinite series using finite-differences. As you can see in the ($*$) equation, the author has achieved the desired issue using the central-differences. The more I struggle to understand how the author has achieved to the aforementioned series, the less I progress! 
It is worth mentioning that the book hasn't have explained about the way any more.
What I've get so far is that the central-differences of second order are written for three nominal. 
Second Order Central Differences
But the weird thing is that the denominator of the third expression is 180, while by using the central-differences it would be 90 in the very text of the book.
Please, someone enlighten me regarding the ($*$) series (how is that produced?!)
 A: Following is an informal description, but it should give you an idea how one discover such a formula.
Let $X$ be a space of test functions on $\mathbb{R}$ which are smooth and sufficiently regular that following descriptions make sense. On $X$, define
an operator $D$ and a family of operators $T_a$, one for each $a \in \mathbb{R}$ as follows:


*

*For any $\phi \in X, x, a \in \mathbb{R}$,
$$(D\phi)(x) = \frac{d\phi}{dx}(x)\quad\text{ and }\quad (T_a\phi)(x) = \phi(x+a)$$


i.e. $D$ is the operator for taking derivative and $T_a$ is the operator to translate the function for an amount $-a$. If we apply Taylor series expansion
to $T_a\phi$, we find
$$(T_a\phi)(x) = \phi(x+a) = \sum_{n=0}^\infty \frac{a^n}{n!} \frac{d^n\phi(x)}{dx^n} = \sum_{n=0}^\infty \frac{a^n}{n!} (D^n\phi)(x)$$
So formally, we have
$$T_a = \sum_{n=0}^\infty \frac{a^n}{n!}D^n = e^{aD}$$
In terms of $T_a$, the central difference operator $\delta_x$ is simply $T_{\frac12} - T_{-\frac12}$. This leads to
$$\delta_x = e^{\frac{D}{2}} - e^{-\frac{D}{2}} = 2\sinh\left(\frac{D}{2}\right)
\quad\implies\quad  D^2 = \left(2\sinh^{-1}\left(\frac{\delta_x}{2}\right)\right)^2\tag{*1}$$
If you throw the function $(2\sinh^{-1}(t/2))^2$ to an CAS and expand it as a power series of $t$,
you will get
$$(2\sinh^{-1}(t/2))^2 =
t^2 - \frac{t^4}{12} + \frac{t^6}{90} - \frac{t^8}{560} + \frac{t^{10}}{3150} - \cdots$$
Substitute this expansion into RHS of $(*1)$ and apply suitable scaling  (because $h \ne 1$), you
obtain the mysterious representation of $\frac{d^2}{dx^2}$ as a power series of $\delta_x^2$. 
About the second half of your question, I have no idea what are 
those expressions after the line
the central-differences of second order are written for three nominal
in your question.
The coefficients for the even-order
higher-order differences I know are all binomial coefficients and forming a row of Pascal's triangle.
