I have started learning lambda calculus from the book by Hindley and Seldin.It brought up the concept with the function $x-y$, first as a function of $x$ and then as a function of $y$. In a way, it emphasized the difference between treating the expression as a function of $x$ and function of $y$, giving separate names to the functions.$$f(x,y) =  x-y \qquad\text{and}\qquad g(y,x) = x -y$$and, in $\lambda$ notion,$$h = \lambda xy.x-y \qquad\text{and}\qquad g = \lambda yx. x-y\,.$$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.But then it said that this can be denoted as, with $h$ being the common name of the function$$h = \lambda xy.(x-y)\qquad\text{and}\qquad h = \lambda yx.(x-y)\,.$$Is it so that in the above expression the variable which is adjacent to $\lambda$ receives the value and the other one is constant?The above function is called a two-place function; what does that mean?Then he introduces the following function in $\lambda$ notation calling it a one-point function:$$h^\star = \lambda x .(\lambda y . xy)$$and says that, for each number $a$, we have $h^\star(a)=\lambda y . ay$.Here $a$ is being provided as an argument but $\lambda$ is placed near $y$. Why is this?And then it deduces that $(h^\star(a))(b)=(\lambda y . ay)(b) = ab = h(a,b)$ and says $h^\star$ can be viewed as representing $h$. Is representing a technical term? How was  $(\lambda y . ay)(b) = ab$ concluded?
Basic notation in lambda calculus
lambda calculus
If I am not wrong then it can be inferred that :$f=\lambda x.(xy)$ 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:$$\begin{align*}plus_z &:= f^\prime(s) = s + z\\ times_y &:= f^\star(s) = s\times y\end{align*}$$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:$$\begin{align*}h^\star &= \lambda x . (\lambda y . x  y)\\h(x,y) &= x-y\end{align*}$$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 $(y.ay)(b)=ab$ concludedThe 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 $h(a,b) = a - b$.  