An object in this category is a pair of sets and a map between them. Write $(A \to B)$ for example.
Usually a morphism $f$ in a category has only one range $Dom(f)$ and codomain $Cod(f)$ assigned to it. In a double category there are two flavours of each for every morphism. A morphism in this category is a commutative square like this. . . .
$A \ \to \ B$
$\downarrow \ \ \ \ \ \ \ \ \downarrow$
$C \ \to \ D$
The 'vertical' domain and codomain are $Dom_{|} = (A \to B)$ and $Cod_| = (C \to D)$
The 'horizontal' domain and codomain are $Dom_-=(A \to C)$ and $Cod_- =(B \to D)$
There are two 'vertical' and 'horizontal' notions of composition , call them $\circ_|$ and $\circ_-$. But we can only compose two morphisms using $\circ_|$ for example if the corresponding vertical domains match up. What this means is the bottom row of the first matches the top row of the second.
We compose vertically by putting the first on top of the second to get a tall rectangle. The corners of the rectangle form a new square. That square is the vertical composition.
Likewise two squares can only be composed horizontally if the right column of the first matches the left column of the second. We compose horizontally by putting the first left of the second to get a long rectangle. The corners of the rectangle form a new square. That square is the horizontal composition.
It can occur that two morphisms may be composable horizontally but not vertically or vice versa.
So why is this not a $2$-category? Well what exactly is an identity morphism with respect to $\circ_-$? It is a morphism $e$ such that $Dom_-(e) = Cod_-(e)$ and $f \circ_- e = e \circ_- f = f$ for every morphism $f$ where all parts are defined.
By experimenting you can show these morphisms are exactly the squares of the form. . .
$A \ \to \ A$
$\downarrow \ \ \ \ \ \ \ \ \downarrow$
$C \ \to \ C$
. . . where the horizontal arrows are the identity map and the vertical arrows match. Meanwhile the vertical identity morphisms are of the form. . . .
$A \ \to \ B$
$\downarrow \ \ \ \ \ \ \ \ \downarrow$
$A \ \to \ B$
. . . where the vertical arrows are the identity map and the horizontal arrows match.
But clearly a square may be of one type but not the other.