Lets allow ourselves to be a bit informal for the sake of exploration. Assume that for any sufficiently nice $f$ in two variables $x$ an $y$, we have $$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$ and that $d$ is linear and satisfies the product rule, by which I mean $d(fg) = d(f)\cdot g+f\cdot d(g).$ From a differential forms perspective we have $d(df) = 0$, but lets just ignore that for now; that is, do not assume $d(df) = 0$. Also, lets treat products as symmetrical rather than antisymmetrical.
We compute:$$\begin{align} d(df) &= d\left(\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \right) \\ & = d\left(\frac{\partial f}{\partial x} dx\right) + d\left(\frac{\partial f}{\partial y} dy \right) \\ & = d\left(\frac{\partial f}{\partial x}\right) dx+ \frac{\partial f}{\partial x} d(dx)+ d\left(\frac{\partial f}{\partial y}\right) dy+ \frac{\partial f}{\partial y} d(dy) \\ & = \left(\frac{\partial }{\partial x}\frac{\partial f}{\partial x} dx + \frac{\partial }{\partial y}\frac{\partial f}{\partial x} dy\right) dx + \frac{\partial f}{\partial x} d(dx) + \left(\frac{\partial }{\partial x}\frac{\partial f}{\partial y} dx + \frac{\partial }{\partial y}\frac{\partial f}{\partial y} dy\right) dy + \frac{\partial f}{\partial y} d(dy) \\ & = \frac{\partial }{\partial x}\frac{\partial f}{\partial x} dx^2 + 2\frac{\partial }{\partial y}\frac{\partial f}{\partial x} dx dy + \frac{\partial }{\partial y}\frac{\partial f}{\partial y} dy^2 + \frac{\partial f}{\partial x} d(dx)+\frac{\partial f}{\partial y} d(dy) \end{align}$$
So if we can just find $d(dx)$ and $d(dy)$, then we've found a formula for $d(df)$ that isn't zero. Indeed, it seems to be the quadratic part of the second-order Taylor expansion for $f$. It's therefore tempting to declare that $d(dx)$ and $d(dy)$ are zero. This feels a bit dodgy though. Let's feel less bad about it by declaring that if all second-order partial derivatives of $f$ are zero, then $d(df) = 0$. This yields $d(dx) = 0$ and $d(dy)=0$ as a special case, as desired, but feels less depressing.
With this stipulation, we find that $$d(df) = \frac{\partial }{\partial x}\frac{\partial f}{\partial x} dx^2 + 2\frac{\partial }{\partial y}\frac{\partial f}{\partial x} dx dy + \frac{\partial }{\partial y}\frac{\partial f}{\partial y} dy^2.$$
More generally, I think we find that $d^n f$ tells us the $n$-th order part of the Taylor series of $f$. (I'm not completely sure about this claim actually - thought anyone?). Anyway, assuming this is true, $e^{d} f$ is basically the whole Taylor series, so long as we perform the shady looking computation $$e^d f = \left(\sum_{i = 0}^\infty \frac{d^i}{i!}\right) f = \sum_{i = 0}^\infty \left(\frac{d^i}{i!}f\right)$$
Question. Does this kind of thing go anywhere? This seems relevant.
Remark. I think we should be able to move this kind of reasoning to arbitrary smooth manifolds, by choosing an atlas and computing $d^n f$ locally. We can then either declare that $d(dx_i)$ is zero for each coordinate function $x_i$, or we can do the second-derivative trick; in either case I suspect we'll get the same result, and my guess is that the answer should be atlas-independent.
Further Remark. If we want, we can set all terms above degree $n$ to zero. In this way, we find that $e^d f$ is telling us the $n$-th order Taylor polynomial of $f$ at each point of our space. The case where we kill all terms above degree $1$ to $0$ yields the familiar formula $d(df) = 0$ from differential topology, just by observing that the above formula for $d(df)$ is a linear combination of terms of degree $2$.
