Relation between exterior (second) derivative $d^2=0$ and second derivative in multi-variable calculus.

Good question! Here's a start. The ordinary derivative in one-variable calculus is a Lie derivative along a special vector field on $\mathbb{R}$; in particular, it is not a special case of the exterior derivative. The exterior derivative is instead some kind of "universal derivative": it records all of the information you would need to determine the derivative of a function along any vector field, for example. In particular, unlike the ordinary derivative, the exterior derivative of a function is a different kind of object, namely a $1$-form. Roughly speaking, a $1$-form is "the kind of thing that pairs with a vector field to return a number," so you can see the relationship there to what I said above.

That $d^2 = 0$ is probably something you were already taught in vector calculus.

For instance, you probably remember that $\nabla \times \nabla \phi = 0$ for $\phi$ a scalar field, or that $\nabla \cdot (\nabla \times E) = 0$ for $E$ a vector field. It's a good exercise to show that both of these can be written as $d^2 f = 0$ and $d^2 E = 0$.

Of course, you know that $\nabla$ can be written in terms of partial derivatives:

$$\nabla = (dx) \frac{\partial}{\partial x} + (dy) \frac{\partial}{\partial y} + dz \frac{\partial}{\partial z}$$

You should also know that the above vector calculus identities rely upon the equality of mixed partial derivatives:

$$\nabla \times \nabla f = \frac{\partial^2 f}{\partial x \partial y} dx \times dy + \frac{\partial^2 f}{\partial y \partial x} dy \times dx + \ldots$$

But since $dx \times dy = -dy \times dx$, the equality of mixed partial derivatives reduces this to zero. The logic is similar for the vector field case.