Real function of one variable 
A real function of one variable is a set $f$ of ordered pairs of real numbers such that for every real number $'a'$ one of the following two things happen:
  (i) There is exactly one real number $b$ for which the ordered pair $(a, b)$ is a member of $f$. In this case $f(a)$ is defined and we write $f(a) = b$ the number $b$ is called the value of $f$ at $a$. 
  (ii) $f(a)$ is undefined if there is no real number $b$ for which the ordered pair $(a,b)$ is a member of $f$. 

If this is the ordered pair $(4, 2)$ $f(4) = 2$, where a = 4 and b = 2. If the value of $b$ becomes 5, then $f(4) ≠ 5$. Is it what the above quoted passage saying? 
 A: $f$ in this case is the set of all points $(x, f(x))$, for $x \in$ the domain on which $f$ is defined. $\quad x\overset{f}{\mapsto } f(x)$
If $x \in X$, $X$ being the domain of $f$, then $f(x)$ maps each $x \in X$ to a real value; the collection of all ordered pairs: $(x, f(x))$ to which $f$ maps $x$ can be considered a set in its own right.
With respect to your added example. If $(4, 2) \in f$, then we would cease to have a function if $(4, 5) \in f$. That is, $f$ would map $4$ to two distinct values.
If there exists no $b = f(a)$, for a given $a$, then the function is not defined at $x = a$: that is, $f(a)$ is undefined.
A: It would help if you realized that the so-called set $f$ is the graph of your function. Let me recall some key facts about such graphs.
Let $G$ be a subset of $\mathbb{R}^2$. Then $G$ is the graph of a function if and only if, for every $a\in\mathbb{R}$, the vertical line $\{(x,y)\;;\; x=a\}$ has at most one intersection point with $G$. Some people call this the vertical line test. 
Let $G$ be such a set. Then we define the corresponding function $f$ as follows. The domain of $f$ is the set of all $a$ for which $G\cap\{x=a\}\neq\emptyset$. For such an $a$, the intersection is a single point $(a,b)$ and we set $f(a):=b$.
Conversely, check that if $f:D\longrightarrow \mathbb{R}$ is a function, then
$$
G:=\{(a,f(a))\;;\;a\in D\}
$$
actually defines a subset of $\mathbb{R}^2$ which has the property stated above. This proves the characterization I started with.
To answer your question: if $(4,2)$ is on the graph of $f$, this means $f(4)=2$. If it is $(4,5)$, then $f(4)=5$. But both can't happen simultaneously, as a function takes only one value at a given point of its domain. That is the intersection characterization with vertical lines I started with.
