To understand this we need to take a step back and consider what we mean by current. We tend to think a current as that which flows in a wire, but more generally if we have some charge moving in some material then we describe this by a current density, $\mathbf J$, and this is a vector field. Then if we choose some area element $d\mathbf A$ the current through this area is given by the dot product:

$$ dI = \mathbf J \cdot d\mathbf A $$

And to get the total current though a surface we just integrate $dI$ over the surface:

$$ I = \int_S \mathbf J \cdot d\mathbf A $$

So showing that the current is scalar is just a matter of showing that the dot product is not changed by a coordinate transformation. For linear transformations this is obvious because the transformation does not change the norms of the vectors nor the angle between them so $|\mathbf J||d\mathbf A|\cos\theta$ is necessarily constant.

To understand this we need to take a step back and consider what we mean by current. We tend to think a current as that which flows in a wire, but more generally if we have some charge moving in some material then we describe this by a current density, $\mathbf J$, and this is a vector field. Then if we choose some area element $d\mathbf A$ the current through this area is given by the dot product:

$$ dI = \mathbf J \cdot d\mathbf A $$

And to get the total current though a surface we just integrate $dI$ over the surface:

$$ I = \int_S \mathbf J \cdot d\mathbf A $$

So showing that the current is scalar is just a matter of showing that the dot product is not changed by a coordinate transformation. For linear transformations this is obvious because the transformation does not change the norms of the vectors nor the angle between them so $|\mathbf J||d\mathbf A|\cos\theta$ is necessarily constant. I confess I'm not sure how you extend this to a more general transformation.