Let $f$ be a function and define $\nabla_X f = X(f)\,\,(1)$, where $\nabla$ is the connection on a manifold and as far as I understand the r.h.s is a function and $X$ is a vector field.
I am just a bit confused with the following statement: 'Since $\nabla_i f = \partial_i f, $ the covariant derivative of the covector $\nabla f$ is $\nabla_i \nabla_j f = \partial_i \partial_j f - \Gamma^k_{ji}\partial_k f$. But why is $\nabla f$ a covector. Isn't it a function from eqn $(1)$? Thanks!
You have two different things here, $ \nabla_X f $ and $ \nabla_i f $. The first has the coordinate-dependent expression $ X^i \nabla_i f $; this means that, as $X(f)$ is a scalar function, $\nabla_i f$ is a linear map from vectors to differentiable scalar functions, i.e. $T(M) \to C^{k}(M)$. Thus it is in the dual space of $T(M)$, $T^*(M)$, and these are covectors. Therefore $\nabla f$ is a covector.
Perhaps the easiest way of seeing this is to note that $X(f)$ is coordinate-independent, so a $(0,0)$-tensor. $X^i$ is a $(1,0)$-tensor, so by the quotient theorem, $\nabla_i f$ must be a $(0,1)$-tensor, i.e. a covector.
I hope that answers your question.