Let $\nabla$ be a connection on a smooth manifold $M$, $X$ a vector field on $M$, and $f$ a real-valued function on $M$. I would like to show that $$\nabla_Xf = Xf.$$
In addition to showing this technically, why is this true geometrically? Covariant derivatives allow us to connect nearby tangent spaces, but what is the interpretation when $f$ is a function and so no tangent spaces are involved?
Actually，we can define the connection in each tensor bundle $T^{(k,l)}TM$ such that we can compute covariant derivatives of tensor fields. The smooth function $f$ can be seen as a $(0,0)$ type tensor field. Indeed, a linear connection is a principle of computing direction derivative. So the $\nabla_X f$ is the derivative of $f$ in the direction $X$.