Let me try to explain it formally. I'll be using the 2-dimensional plane \mathbf R^2, but you can easily generalize this with \mathbf R^n.
In affine geometry, we can see \mathbf R^2 as a set :
- of points (let's note simply \mathbf R^2 the set \mathbf R^2 seen as a set of points)
- or a vector space, in other words, a set of "vectors" (let's note \overrightarrow{\mathbf R^2} the set \mathbf R^2 seen as a set of vectors).
Note that \mathbf R^2 and \overrightarrow{\mathbf R^2} are the
same set (the set of couples of reals), but perceived differently.
You can make linear combinations (sum, difference, multiplication by a scalar) in \overrightarrow{\mathbf R^2}, using the usual operations on vectors. More formally, we define + as the binary operation (a,b) + (c, d) = (a+c, b+d), and so on.
Note that you can only do this with \overrightarrow{\mathbf R^2} : as, for instance, adding two vectors seems pretty intuitive (you just put the start of the second arrow at the end of the first one), it's counterintuitive to visualize the same operation done with points (if you want to visualize it with points, you'd actually "imagine" vectors between the origin and the points).
If we take two points A and B in \mathbf R^2 (notice there's no arrow here, A and B are points) we
define the
vector \overrightarrow{AB} as the difference B - A. You might ask yourself: "but I thought we couldn't make sums, differences, etc. with points as we could with vectors". Well yes, but here we'll just use the regular definition of the substraction of two couples of reals, i.e. if A, B have (x_A, y_A), (x_B, y_B) for standard coordinates, we'll define B-A = (x_B - x_A, y_B - y_A).
Now, if you take four points, A, B, A', B', such that "the arrow that links A and B is equal to the arrow that links A' and B'", it is easy to show that the difference B-A will be equal to the difference B'-A', in other words, that \overrightarrow{AB} = \overrightarrow{A'B'}.