The exterior derivative of a zero-form is a one-form
$ df = \frac{\partial f}{\partial x} e^{x} + \frac{\partial f}{\partial y} e^{y} + \frac{\partial f}{\partial z} e^{z} = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz $
The exterior derivative of a one-form is a two-form
$dF = \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \mathbf{e^{xy}} + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \mathbf{e^{zx}} + \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \mathbf{e^{yz}}$
df is the total derivative of f so it makes perfect sense to call it df.
But dF is not the total derivative of F. So why call it dF?