Covariant derivative of a function confusion So the covariant derivative of a scalar function $f$ on a manifold $M$ w.r.t the vector $X$ is defined as 
$$ \nabla_X f = X(f) .$$
From the very beginning of my course on general relativity, it has been stated that vectors in the tangent space are directional derivative operators. So $X$ is a map from the space of functions on the manifold $\mathcal{F}$ to the reals:
$$ X : \mathcal{F} \rightarrow \mathbf{R}. $$
However, I am fully aware that $\nabla_X f = X(f)$ defines a $(0,1)$ tensor, but this seems to contradict the definition that $X(f)$ is a scalar? 
 A: It depends on how you read the symbol
$$\nabla_X f.$$
If you consider $X$ and $f$ fixed, i.e., you pick one specific vector field $X\in \Gamma(TM)$ and a specific smooth function $f\in C^\infty(M)$, then $\nabla_X f$ is just the specific function $X(f)$.
On the other hand, if you consider $f\in C^\infty(M)$ to be a fixed function and let $X$ be an arbitrary vector field, so that you are actually looking to the map $X\mapsto \nabla_X f$, then you have a $(0,1)$ tensor field usually denoted as $\nabla f$, whose action on $X$ is $\nabla f(X) = \nabla_X f$.
Why? Because of the properties of the covariant derivative. It is well known that $\nabla_X Y$ is defined to be $C^\infty(M)$-linear (or tensorial as some people prefer calling it) on the entry below.
This ensures that fixing $f$ you get $X\mapsto \nabla_X f$ a $C^\infty(M)$-linear mapping defined on vector fields, and hence a $(0,1)$ tensor field.
The thing is just that in one case $\nabla_X f$ is one specific calculation with a specific result and the other is actually one mapping, i.e., a function. This is the same as asking whether $f(x)$ is a function or a number. Some people read this with $x$ one arbitrary variable so that $f(x)$ is the "rule that defines $f$" and so is the function, but if $x\in \mathbb{R}$ is one specific number $f(x)$ is a number as well.
A: On a smooth manifold $M$, the tangent vector fields $v\in \Gamma_M(TM)$ may be defined as a collection of paths $\gamma_x : \mathbb R \to M$ with $\gamma(0) = x$ for all points $x\in M$, being equivalent if the derivative $v_x$ coincides.
If we have a scalar function $f : M \to \mathbb R$ defined on the manifold, we can naturally find a map from $\mathbb R$ to $\mathbb R$ by defining $f \circ\gamma _x$. Then the rate of change of $f$ along $\gamma$ at $x$ is given by,
$$(D_vf) (x):= \mathrm d(f \circ \gamma)_0.$$
This motivates a derivation $D_v : C^\infty(M) \to C^\infty(M)$ which we know as $v^\mu\partial_\mu$ when acting on a scalar and it can be proven every derivation can be shown to arise from a tangent vector field. We can also interpret $D_v$ as the Lie derivative along $v$ for a scalar.
Notice that explicitly in components,
$$v^\mu \nabla_\mu f = v^\mu \partial_\mu f = \sum_{i=1}^d v^i \partial_i f$$
which is clearly a scalar, since there are no free indices left; we have only one pair which have been contracted giving the sum. If on the other hand we write $\nabla_\mu f$, this is a $(0,1)$ tensor since we have a free index $\mu$, uncontracted. $\nabla_\mu f$ is just the covariant derivative then, unrelated to any vector field.

Explicit Example
Take a vector field $V^\mu$ and a function $\phi$ on four-dimensional space time, endowed with Cartesian coordinates. Then we have that,
$$\mathcal L_V \phi = V^\mu \nabla_\mu \phi = V^\mu \partial_\mu \phi = V^t\frac{\partial}{\partial t}\phi + V^x\frac{\partial}{\partial x}\phi+V^y\frac{\partial}{\partial y}\phi+V^z\frac{\partial}{\partial z}\phi$$
which is clearly a scalar since components of a vector are scalar, and derivatives of scalars are scalar. On the other hand,
$$\nabla_\mu \phi = \partial_\mu \phi = \begin{pmatrix}
\partial_t\phi \\ 
\partial_x\phi \\ 
\partial_y \phi \\ 
\partial_z \phi
\end{pmatrix}^T$$
which is clearly a $(0,1)$ tensor.
