Group action and orbit space
Here is my best attempt to answer your question,
When thinking of the set of all orbits, what comes to mind for instance in $\mathbb{R}^2$, is loads of lines/curves. However if I consider $X/G$, where each orbit of $X$ is an equivalence class, is each orbit not represented by a point?
Let $X$ be a set equipped with an action of a group $G$. There are several different ways to think about the orbit space, $X/G$. Let's use some of my favourite examples to illustrate:
Example 1 Let $S^1 = \{ w\in \mathbb{C} \mid |w| = 1\}$ be the group of unit complex numbers (the group operation is multiplication of complex numbers). Let $S^2$ be the sphere of unit vectors in $\mathbb{R}^3$. Let $S^1$ act on $S^2$ by rotation around the $z$-axis in $\mathbb{R}^3$, by the formula, $$ w \cdot (x,y,z) = (\cos(\theta)x,\sin(\theta)y,z), \quad w = e^{i\theta}. $$ The orbits of this action are the circles on $S^2$ that are circles of latitude (equal to the intersection of the sphere with an affine plane orthogonal to the $z$-axis) and two points: the north pole $(0,0,1)$ and the south pole, $(0,0,-1)$. Here is a nice picture:
Example 2 Let the multiplicative group $\mathbb{R}^{\times}$ act on $\mathbb{R}^2\setminus \{(0,0)\}$ by scaling: $$ t \cdot (x,y) = (tx,ty). $$ The orbits of this action are lines through the origin in $\mathbb{R}^2\setminus \{(0,0)\}$. Each line $y = mx$ is an orbit, along with the vertical line, $x=0$.
Example 3 This might take longer to chew on. Let $\mathbb{Z}$ be the group of integers with addition. Let $S^1$ be the set of unit complex numbers (considered only as a set, not a group). For any real number $\beta \in \mathbb{R}$, we can define an action of $\mathbb{Z}$ on $S^1$ using multiplication of complex numbers: $$ k\cdot w = e^{2\pi i \beta k}w. $$ If $\beta$ is a non-zero rational number of the form $\frac{p}{q}$ with $gcd(p,q)= 1$, then each orbit is a set of $q$ points. If $\beta$ is irrational, then each orbit contains infinitely many points and is dense in $S^1$.
Alright, back to the general setting, where $X$ is a set and $G$ is a group. By definition, $X/G$ is the set of orbits of the action of $G$. This is understandably confusing because
an element in $X/G$ is a set of elements of $X$.
First, let's picture this correctly for example 1. Each of the black circles in the image is an element of $S^2/S^1$. It is a set of points in $S^2/S^1$.
The small black dot at the north pole (and the small black dot at the south pole, which you can't see because of perspective) is also an element of $S^2/S^1$. For example 2, each of the lines $y=mx$ is an element of $\mathbb{R}^2\setminus \{(0,0)\}/\mathbb{R}^{\times}$. For example 3, when $\beta$ is irrational, it is more difficult to visualize.
Second, let's try to clear up a subtlety in the meaning of the words "point" and "element." The word "element" is very general. Whenever we have a set, objects that are contained in the set are called elements. For example, an orbit $G\cdot x$ is an element of the set $G/X$. In contrast, when we use the word "point" we often imply (intentionally or not) a more geometric setting. For instance, it is natural to call $(1,0)$ a point in $\mathbb{R}^2$, but it is completely psychotic to call the line $y=mx$ a point. It is a line! It is also an element of the orbit space $\mathbb{R}^2\setminus \{(0,0)\}/\mathbb{R}^{\times}$.
Now let me completely contradict the previous paragraph. There is little or no difference between the words "point" and "element." Before trying to explain, let me give an example of a sentence that makes perfect sense but seems to contradict what I said in the previous paragraph.
Let $P^1$ be the space of lines through the origin in $\mathbb{R}^2$. For example, $y = 2x$ is a point in $P^1$.
The important thing here is the following: $y = 2x$ is not a point in $\mathbb{R}^2$, but $y = 2x$ is a point in $P^1$.
What we really mean is that $y = 2x$ is an element of $P^1$, but we often think of the space $P^1$ geometrically, so we often use the word point instead of element. As long as you understand which set an object is an element of, it's perfectly fine to say "point" instead of "element." You will find that mathematicians do this constantly without thinking about it.
In fact, thinking about the space $X/G$ more geometrically -- thinking of elements of this set as points in a space $X/G$ rather than sets of points in $X$ -- is an important shift in perspective. This shift in perspective often comes with a more detailed study of $X/G$.
In many settings where we consider the action of a group $G$ on a set $X$, the set $X$ has additional structure (such as a metric, or a topology). If the action of $G$ preserves this structure, then the set $X/G$ often (but not always) inherits a structure of the same kind. So the orbit space often becomes more than just a set, and one often wants to study the extra structure on $X/G$ in a geometric or topological way.
For example, consider $S^1$ acting on $S^2$. Recall that there is a notion of distance (aka a metric) on $S^2$ (the distance between two points is the length of the shortest geodesic arc between them). The action of $S^1$ preserves this metric (since it sends geodesic arcs to other geodesic arcs). The orbit space $S^2/S^1$ inherits a notion of distance that can be defined as follows: the distance between two orbits equals the shortest distance between two points contained in each orbit (if you know the definition of a metric this an exercise to check the definition).
This doesn't always work. For instance, in Example 3 the circle has a metric and the action of $\mathbb{Z}$ preserves it. But this doesn't define a metric on $S^1/\mathbb{Z}$ when $\beta$ is irrational since the distance between every orbit would be 0.
One last comment, relating to your question.
An alternative way to understand the space $X/G$ is to identify it with a subset of $X$ that intersects every orbit exactly once. Such a subset is often called a "section" for the action. Take Example 1 again.
Consider an arc, $A$, in $S^2$ that consists of half a longitudinal circle, from the south pole to the north pole (but not the way back on the other side). More explicitly, we can take $$ A = \{ (0,0,\cos(\theta)) \mid 0 \leq \theta \leq \pi\}.$$
This arc intersects every orbit of the $S^1$ actin exactly once. Thus you can define a map
$$ S^2 /S^1 \rightarrow A, \quad S^1\cdot x \mapsto a $$
where $a\in A$ is the unique element in the intersection $(S^1\cdot x) \cap A$. This map is a bijection. Moreover, if you consider $S^2/S^1$ with the metric defined above, and $A$ with the metric inherited from $S^2$ as a subspace, then this map is an isomorphism of metric spaces (exercise).
Of course, you can pick different sections for this example.