He doesn’t explain anything there in the question because he’s already explained it in the text, probably with examples.
Let’s forget the summation notation for now and concentrate on the ideas. You need to divide the interval $[1,3]$ into $n$ equal subintervals. Since $[1,3]$ is $3-1=2$ units long, each of the $n$ subintervals must be $\frac2n$ units long. For instance, if $n$ were $5$, you’d divide the interval into subintervals of length $\frac25=0.4$: $[1.0,1.4],[1.4,1.8],[1.8,2.2],[2.2,2.6]$, and $[2.6,3.0]$. $\Delta x$ denotes the length of the subintervals, so in this problem $\Delta x=\frac2n$. (In my numerical example $\Delta x=\frac25$.)
The numbers $x_0,x_1,x_2,\dots,x_n$ are the endpoints of the subintervals. The first subinterval is $[x_0,x_1]$; the second is $[x_1,x_2]$; the $n$-th and last is $[x_{n-1},x_n]$; and in general the $k$-th subinterval is $[x_{k-1},x_k]$. Of course this means that $x_0=1$ and $x_n=3$, since the first interval must begin at $1$, and the last must end at $3$. So how do we find the other numbers $x_k$?
Remember that each interval $[x_{k-1},x_k]$ is $\Delta x=\frac2n$ units wide. Thus, $x_1$ is $\frac2n$ units to the right of $1$, $x_2$ is $\frac2n$ units to the right of $x_1$, and so on. This means that
$\begin{align*} x_0&=1\\ x_1&=1+\frac2n\\ x_2&=1+\frac2n+\frac2n=1+2\left(\frac2n\right)\\ x_3&=1+2\left(\frac2n\right)+\frac2n=1+3\left(\frac2n\right)\\ x_4&=1+3\left(\frac2n\right)+\frac2n=1+4\left(\frac2n\right) \end{align*}$
and so on. In general $x_k=1+k\left(\frac2n\right)=1+\frac{2k}n\;.$
Now we can start to plug some specifics into the expression
$f(x_1)\Delta x+f(x_2)\Delta x+\dots+f(x_n)\Delta x\;:$
it’s $f\left(1+\frac2n\right)\left(\frac2n\right)+f\left(1+\frac4n\right)\left(\frac2n\right)+f\left(1+\frac6n\right)\left(\frac2n\right)+\dots+f\left(1+\frac{2n}n\right)\left(\frac2n\right)\;,$ where I’ve simply plugged in the values that we found above for $x_1,x_2,\dots,x_n$ and $\Delta x$. There’s one term for each of the $n$ subintervals, and the term for the $k$-th subinterval is $f(x_k)\Delta x=f\left(1+\frac{2k}n\right)\left(\frac2n\right)\;.$ Since $k$ runs from $1$ through $n$ (first interval through $n$-th and last interval), that rather large and unwieldy sum can be abbreviated to $\sum_{k=1}^nf\left(1+\frac{2k}n\right)\left(\frac2n\right)\;.$ Every term has a factor of $\frac2n$, so I can factor it out, if I wish, to get $\frac2n\sum_{k=1}^nf\left(1+\frac{2k}n\right)\;.\tag{1}$
We’re not done, though, because we actually know what the function $f$ is: $f(x)=\frac{2x}{x^2+1}\;.$ Inserting that into $(1)$ gives us $\frac2n\sum_{k=1}^n\frac{2\left(1+\frac{2k}n\right)}{\left(1+\frac{2k}n\right)^2+1}\;.$
Now $\frac{2\left(1+\frac{2k}n\right)}{\left(1+\frac{2k}n\right)^2+1}$ looks pretty horrendous, but it can be simplified quite a bit:
$\begin{align*} \frac{2\left(1+\frac{2k}n\right)}{\left(1+\frac{2k}n\right)^2+1}&=\frac{2+\frac{4k}n}{1+\frac{4k}n+\frac{4k^2}{n^2}+1}\\ &=\frac{\frac{2n+4k}n}{\frac{2n^2+4kn+4k^2}{n^2}}\\ &=\frac{n(2n+4k)}{2n^2+4kn+4k^2}\\ &=\frac{n(n+2k)}{n^2+2kn+2k^2}\;, \end{align*}$
so $\frac2n\sum_{k=1}^n\frac{2\left(1+\frac{2k}n\right)}{\left(1+\frac{2k}n\right)^2+1}=\frac2n\sum_{k=1}^n\frac{n(n+2k)}{n^2+2kn+2k^2}=2\sum_{k=1}^n\frac{n+2k}{n^2+2kn+2k^2}\;.$ Finally(!), we want the limit of this sum as $n\to\infty$:
$A=\lim_{n\to\infty}2\sum_{k=1}^n\frac{n+2k}{n^2+2kn+2k^2}=2\lim_{n\to\infty}\sum_{k=1}^n\frac{n+2k}{n^2+2kn+2k^2}\;.$