About covariant derivative on scalar functions My question is . When we discuss covariant derivative on any tensorial object we say we apply an operator which reduces to ordinary partial derivative while acting on scalar function. Why it reduces to ordinary partial derivative?
 A: The covariant derivative can be thought of as a generalization of the ordinary derivative in flat space, which takes into account the fact that tangent spaces at different points are generally not naturally identified because of curvature.
A scalar function is not dependent on this identification of the tangent spaces, and so the covariant derivative just reduces to the ordinary derivative when acting on them.
A: From an operational point of view, the need for a covariant derivative arises when we attach distinct copies of some vector space $F$ to each point of a "base space" $M$.  The standard example which arises in GR is the attachment of the tangent space to each point of the underlying spacetime $M$, in which case $F= \mathbb R^{d+1}$.
When we want to take the derivative of a function, we normally think of a difference quotient:
$$ s' (x):= \lim_{\epsilon\rightarrow 0} \frac{s(x+\epsilon) - s(x)}{\epsilon}$$
However, this doesn't work here because $s(x+\epsilon)$ is a vector which lives in the copy of $F$ attached to the point $x+\epsilon$, whereas $s(x)$ lives in the copy of $F$ attached to the point $x$.  Since these are vectors which live in different vector spaces, their subtraction is not well-defined.
In this sense, $s$ is not a vector-valued function from $M\rightarrow \mathbb R^{d+1}$. The outputs of such a function would all live in the same space, and covariant differentiation would not be necessary. When we attach a distinct copy of $F$ to each point of $M$, we have constructed a so-called fiber bundle, and $s$ is called a section of that bundle.
To resolve this issue, we need to define a "transporter" $T_{x,y}$ - a smooth, linear map which takes a vector which lives at $x$ to a vector which lives at $y$.  Once we've done that, we may define
$$Ds(x) := \lim_{\epsilon\rightarrow 0} \frac{T_{x+\epsilon,x} s(x+\epsilon) - s(x)}{\epsilon}$$
Now both terms in the numerator are vectors which live at the point $x$, and so the derivative is well-defined (though of course, it requires us to provide extra data, namely to specify the action of $T$).  This is the so-called covariant derivative.
However, if we are simply dealing with a scalar function $f:M\rightarrow \mathbb R$ (not a section of some bundle, but a genuine function), none of this is necessary. Because $f(x+\epsilon)$ and $f(x)$ both live in the same copy of $\mathbb R$, we can subtract them without issue, and the ordinary notion of differentiation works as usual.
