I've been trying to make sense of this for days now, but I can't seem to be able to!
How does this
Let $t$ be a parameter equal to the distance from the beginning of the polygonal line till the current point along this line. Let $l_{j_{1}}$ and $l_{j_{2}}$ be values of the parameter defining the beginning and the end of the analyzed part of the polygonal line. The horizontal line approximating a given part of the polygonal line can be defined as $$y=\frac{V_{y}}{l_{j_{2}}-{l_{j_{1}}}}, V_{y}=\int_{{l_{j_{1}}}}^{{l_{j_{2}}}} y(t)dt$$ The vertical line can be found similarly $$x=\frac{V_{x}}{l_{j_{2}}-{l_{j_{1}}}}, V_{x}=\int_{{l_{j_{1}}}}^{{l_{j_{2}}}} x(t)dt$$
Make these straight lines ($L_{i}$) 
What should I plug in as $t$ ? Should I plug in the last point's length?
Are you familiar with the concept of the "average value of a function"?
For a function $f(x)$, the "average vale of $f$ on the interval $[a,b]$" is defined as $$ \mathrm{avg} = \frac{1}{b-a} \, \int_a^b f(x) \, dx $$
If you evaluate this integral, you get a number (a constant), let's call it $c$. Then graphing the horizontal line $y=c$ will show you the average height of the graph between $x=a$ and $x=b$.
This is what is being described in the quote from your post. You have a parameterized curve $r(t) = (x(t), \, y(t))$. Then the quantities: $$ \frac{1}{j_2-j_1} \; \int_{j_1}^{j_2} x(t) \, dt \hspace{0.5cm} \text{ and } \hspace{0.5cm} \frac{1}{j_2-j_1} \; \int_{j_1}^{j_2} y(t) \, dt $$ are just the average $x$-value and average $y$-value of all the points along the curve between times $t=j_1$ and $t=j_2$. In the picture the horizontal and vertical lines are just plotting the equations $x=\mathrm{const}$ and $y = \mathrm{const}$, where "const" is the value of the appropriate integral.