Disclaimer: I know that my question has probably answers in textbooks, but I have only found the answer for quiver representations, and not for modules over quiver algebras. I know that these two notions are closely related, but I could not answer the question posted below.
I have seen the notation of "writing numbers on top of each other" when describing modules over quiver algebras, but (appearently) I do not understand it properly.
My question is: how to obtain this "notation" for a given module?
In particular, the following is from an exercise from class (NOT an exam, we were given a written solution already):
Suppose that we have the following quiver $\Gamma$:
and an admissible ideal $I$ defined by the relations $\alpha_1\alpha_2, \alpha_2\alpha_3, \beta_2\beta_1, \beta_3\beta_2, \beta_1\alpha_1 - \alpha_2\beta_2, \beta_2\alpha_2 - \alpha_3\beta_3, \beta_3\alpha_3$. We have to describe the minimal projective resolutions of all simple modules.
Now the solution given to us begins by claiming that if $A = K\Gamma/I$, then 
Why is the above statement true?
Your notation suggests you are using right modules. Thus the indecomposable projectives are $e_iA$, where $e_i$ is the idempotent corresponding to vertex $i$. We now find a $K$-basis for each of these modules.
$i=1$. We have the paths $e_i$, $\alpha_1$, $\alpha_1\beta_1$. All other paths vanish, since $\alpha_1\beta_1\alpha_1=\alpha_1\alpha_2\beta_2$, and $\alpha_1\alpha_2=0$.
Looking at their end points, these are $1$, $2$, and $1$ respectively. So when drawing the module, we have a two dimensional vector space at vertex 1, with basis $e_1$, $\alpha_1\beta_1$, and a one dimensional vector space at vertex 2, with basis $\alpha_1$. The linear maps come from the natural action of the arrows on this basis, and we see that the module is uniserial, with simple factors 1,2,1 from the top.
$i=2$. We have the paths $e_2$, $\alpha_2$, $\beta_1$ and $\beta_1\alpha_1$ (which equals $\alpha_2\beta_2$).
Their end points are 2,3,1,2 respectively, so the module has a one dimensional vector space at vertex 1, with basis $\beta_1$, a two dimensional vector space at vertex 2, with basis $e_2$ and $\beta_1\alpha_1$, and a one dimensional vector space at vertex 3, with basis $\alpha_2$. Again the linear maps come from the natural action of the arrows on this basis, noting that $\alpha_2\beta_2=\beta_1\alpha_1$.
Here we have simple top 2, the radical has semisimple top 1,3, and the socle is the simple 2 again.
$i=3$. We have $e_3$, $\alpha_3$, $\beta_2$, $\beta_2\alpha_2$.
We have simple top 3, radical having semisimple top 2,4, and simple socle 3.
$i=4$. We have $e_4$ and $\beta_3$. This has simple top 4 and simple socle 3.
Note that these bases combine to give a $K$-basis for $A$, so $\dim A=13$.