Basic notation in lambda calculus
If I am not wrong then it can be inferred that :
  
  $f=\lambda x.(x−y)$ means that while computing the value of the expression the
  expression will vary only with "x" and similarly when "lambda " is
  placed with x the other variable y" will be kept as a constant .


Yes, intuitively that's how it works. 



What you're looking is called Currying and it's explained fairly early in this book (in fact, I think it's the next section but I don't have a copy now to check). So when we want to work with functions with many arguments we often do it like this:

$$ f(x,y,z) = x\times y + z$$

But you can break this into simpler functions, $(x \times y)$ and $(+z)$. The multiplication can be broken into even simpler terms, first I provide $x$, then just multiply by $y$, that is, $f^{\star}(y) = x\times y$. 

Defining some 1-argument functions:

$$
\begin{align*}
plus_z &amp;:= f^\prime(s) = s + z\\
 times_y &amp;:= f^\star(s) = s\times y
\end{align*}
$$

Combining these we can rewrite $f$ as:

$$ f(x,y,z) = f^\prime(f^\star(x)) = plus_z (times_y ~ x) $$ 

Writing this in using $\lambda$ we have:

$$ \lambda x y z. (x\times y + z)  \equiv \lambda x.\bigg(\lambda y. \Big(\lambda z. (x\times y + z ) \Big) \bigg)$$

What the author has shown is that you can take a function with many arguments and transform it into many simpler functions with just one argument, the bigger function is just a composition of the simpler ones. 

So just how the proof works? In your book the following functions are defined:

$$
\begin{align*}
h^\star &amp;= \lambda x . (\lambda y . x − y)\\
h(x,y) &amp;= x-y
\end{align*}
$$

You want to show that $ h^\star = h$. To do that just take any two arguments $a$ and $b$. First evaluate $h^\star$ in $a$.

$$ h^\star(a) = (\lambda x. (\lambda y. x - y))(a) = \lambda y. a - y$$

This gives you a simpler one argument function $h^\prime = h^\star(a)$.


  How was $(λy.a−y)(b)=a−b$ concluded


The same way $(\lambda x. (\lambda y. x - y))(a) = \lambda y. a - y$ was concluded. Just evaluate the function in $b$. 

$$ h^\prime(b) = (h^\star(a))(b) = (\lambda y. a - y)(b) = a - b$$

Which is the same result computed by $h(a,b) = a - b$.