Maybe it's an odd question caused by some overinterpretation..anyway it's not uncommon encounter in books statements like this one (regarding differences among Exterior Derivative, Lie Derivative and Covariant derivative):
All three agree on smooth functions.
from https://math.stackexchange.com/a/264422/277873.
But if $f$ is a smooth function its second order Exterior Derivative $d(df)=0$, which is not true for its second order Lie Derivative $L^2_Xf$...so where am I wrong?
The first order derivative in all 3 cases is a 1-form, which means a pointwise linear function of tangent vectors. In particular, the value of $D_vf(x) = \nabla_vf(x) = \mathcal{L}_vf(x)$ depends only on the value of $v$ at $x$ and not on any of its derivatives. The situation changes when you take the differential, covariant derivative, or Lie derivative of a $1$-form. The value of a Lie derivative of a $1$-form with respect to a vector field $v$ at $x$ depends on not only the value of $v$ at $x$ but also its first derivative. The second covariant derivative of a function, if properly defined (it requires a little care), is a $2$-tensor $\nabla^2 f$, which means it is a pointwise bilinear function of two vectors. In particular, the value of $\nabla^2_{vw}f(x)$ depends only on the values of $v$ and $w$ at $x$ and not on any of their derivatives. If the connection is torsion free, as it is for the Levi-Civita connection, then it is a symmetric $2$-tensor. The exterior derivative of a $1$-form is the antisymmetrization of its covariant derivative (assuming the connection to be torsion free). Since the second covariant derivative of a function is symmetric, it follows that $d(df) = 0$.