One way to think of this is to realise that in $D_{2n}$, you have two kinds of objects that interact - the rotations $r$ and the reflections $s$.
The defining property of the rotation is that $r^n = 1$, that is, after $n$ rotations, you're back to identity.
Similarly, the defining properly of a reflection is $s^2 = 1$, which tells you that two reflections are the same as no reflection.
The next component you need is to relate rotations and reflections:
$$
rs = sr^{-1}
$$
that is, a rotation $r$ followed by a reflection $s$ is the same as the reflection $s$ followed by ($360^\circ$ - $r$).
Now, let us assume that we are given some element of the dihedral group
$$
d = r_0^{i_0}s_0^{j_0} \cdot r_1^{i_1}s_1^{j_1} \ldots
$$
we can issue the requirement that $i_k < n$, since all $i_k$ are the coefficients of the rotations $r_k$, and we can collapse large rotation coefficients to become less than $n$ by using $r^n = 1$
Similarly, we can require that $j_k = 0, 1$ since $s^2 = 1$
Now, to reduce $d$, we can use the fact that $rs = sr^{-1}$ like so:
$$
d = r_0^{i_0}s_0^{j_0} \cdot r_1^{i_1}s_1^{j_1} \ldots
\\ \text {swizzling $r_0$, $s_0$} \\ \\
d = s_0^{j_0}r_0^{-i_0} \cdot r_1^{i_1}s_1^{j_1} \ldots \\
d = s_0^{j_0}r^{i_1 - i_0}s_1^{j_1} \ldots
\\ \text{swizzling $r$, $s_0$ } \\
d = r^{-(i_1 - i_0)}s_0^{j_0}s_1^{j_1} \ldots \\
d = r^{i_0 - i_1}s^{j_0 + j_1}
$$
If that's not clear, we first took $r_0^{i_0} s_0^{j_0} \to s_0^{j_0} r_0^{-i_0}$.
We then composed $r_0^{-i_0} r_1^{i_1} \to r^{i_1 - i_0}$, which gave us one rotation.
Next, we chose to swizzle around $s_0^{j_0}r^{i_1 - i_0} \to r^{-(i_1 - i_0)}s_0^{j_0}$, so we could compose the $s_0$ and $s_1$ together in the next step.
Finally, we compose $s_0^{j_0} s_1^{j_1} \to s^{j_0 + j_1}$.
Clearly, we have reduced a 2-term $r_0^{i_0}s_0^{j_0} r_1^{i_1} s_1 ^{j_1} \to r^{i_0 - i_1} s^{j_0 + j_1}$
This process can be extended to $n$ terms, allowing us to collapse a large presentation of a single element into one $d = r^x s^y$.
Can any relation between elements of the generators be determined from the relations in the presentation?
Yes! Sometimes, a group is defined by one of its presentations, so there's no other choice when that happens.
My inclination is no; because a few lines below, Dummit and Foote say
...in an arbitrary presentation it may be extremely difficult (or even impossible) to tell when two elements of the group (expressed in terms of the given generators) are equal.
This is known as a decidability result. It's the word problem. Generally speaking, given two elements of a group given by a presentation, it is literally impossible to decide whether or not one is equal to another using a Turing machine in finite time.
But this does not mean that any and all relations between elements of a group cannot be determined by the relations of a presentation of that group.
Best Answer
There's some subtlety here, and it's not at all an obvious fact. I'll walk you through the ideas involved, but I'll leave it up to you to verify some details.
Remember what a presentation means. When I write
$$D_8 = \langle r,s ~|~ r^4 = 1, s^2 = 1, srs = r^{-1} \rangle$$
what I mean is that $D_8 \cong F_2 / N$ where $F_2$ is the free group with generators $r$ and $s$, and $N$ is the "normal closure" of $\{r^4, s^2, srsr\}$. That is, $N$ is the smallest normal subgroup containing those elements.
When we quotient, we are imposing relations on the free group. By this I mean we are forcing certain products to be $1$. Why should all of these relations be written as a product of the relations $r^4 = 1$, $s^2 = 1$, and $srsr = 1$? Well for that we need to understand what the normal closure of a set is.
Definition:
Theorem (exercise):
Now we see why the claim is true. Since there are no relations in $F_2$, all relations in $D_8$ come from $N$. But all the relations in $N$ are products of conjugates of the three listed relations. This is the sense in which those relations generate all of them. Of course, there is nothing special about $D_8$. This idea works for any presentation of a group.
I hope this helps ^_^