A real affine space is a triple $(\mathbb A, \varphi, \mathbb V)$ where:
$\mathbb A$ is a set;
$\mathbb V$ is a $\mathbb R$-vector space;
$\varphi: \mathbb A\times \mathbb A\rightarrow \mathbb V$, $(A, B)\mapsto \overrightarrow{AB}$, is a map
satisfying:
$(a)$ (Chasles Identity) $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$;
$(b)$ For every $\overrightarrow{v}\in \mathbb V$ and every $A\in \mathbb A$ there is a unique $B\in\mathbb A$ such that $\overrightarrow{v}=\overrightarrow{AB}$.
From this definition, how to decide when two vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are equal?
From elementary geometry, $\overrightarrow{AB}$ and $\overrightarrow{CD}$ should be equal when we can form a parallelogram using $A$, $B$, $C$ and $D$, but I don't know how to interpret it in the light of the above definition.
Maybe, what is really relevant after all is the equivalence
$$\overrightarrow{AB}=\overrightarrow{CD}\iff \overrightarrow{AC}=\overrightarrow{BD}? $$
Thanks.
I'm not sure that the following will answear your question but I hope it will help.
I think that the goal of affine geometry is kinda the opposite of the fact you want. Indeed, the idea is that you start with a set $\mathbb{A}$ and, if you manage to find a vector space and a map with the properties you listed, then you can think geometrically of the elements $\mathbb{A}$. So, in particular, is the choice of $\varphi$ that gives geometric structure to $\mathbb{A}$.
The classical example of affine space is the set $\mathbb{A}$ of solutions to a non homogeneous linear ODE. In this setting, you can't see a parallelogram looking at the poins of $\mathbb{A}$, they are functions. But, when you add the affine space structure, then you obtain a criterion (the one you wrote) to say that some elements in $\mathbb{A}$ form a parallelogram