It sort of depends on what you mean by substring. Look at the string ABABAB. How many substrings of length $2$ does it have? Start at the first letter. You get AB. Start at the second letter. You get BA. Start at the third letter. You get AB. Start at the fourth letter. You get BA. Start at the fifth letter. You get AB.
There were $5$ places to start, so in a certain sense we got five substrings. But there were only two different ones. So do we say $5$ or do we say $2$? Let us decide it is $5$. Of course, there cannot possibly be any ambiguity if the letters are all different.
So for simplicity assume the letters are all different. Now let's work out a few concrete case, and then everything will be obvious to you.
How many substrings of length $3$ are there in the string $12345678$? We can start anywhere up to and including $6$. So there are $6$ substrings of length $3$. Note that $6=8-3+1$.
How many substrings of length $3$ are there in the string $123456789ABC$? Note that our word has $12$ symbols, and we can start our substring at anything from $1$ to $A$, inclusive. So how many start points are there? Clearly $10$. Note that $10=12-3+1$.
Now we have the string $a_1a_2a_3\cdots a_{47}$. How many substrings of length $3$? The latest we can start is at $a_{45}$, so there are $45$ substrings. Note that $45=47-3+1$.
Now look at the string $a_1a_2a_3\cdots a_{100}$. How many substrings of length $8$? The latest we can start is at $a_{93}$, so there are $93$ substrings, and $93=100-8+1$.
Why is $a_{93}$ the last place we can start? We want to start at a place $p$ such that there are $8$ numbers from $p$ to $100$, inclusive. So there should be $100-8$ numbers from $1$ to $p-1$. Thus $p-1=92$, and therefore $p=93$.
Play with a few more concrete cases, they need not be too big. (So far, I have done all the playing, you have been reading.) Soon everything will be clear.