Partial derivative on manifold In Loring Tu An Introduction to Manifolds pg. 67,  Proposition 6.22: 
I don't understand why $$\frac{\partial r^i}{\partial r^j}(\phi (p))=\delta^i_j.$$
It seems that the author is using $x^i$ and $r^i $ as functions in the numerator and as coordinates in the denominator. Previously the author defined the function $x^i$ as
$$x^i=r^i \circ\phi.$$
Using this definition, the numerator $r^i(\phi(p))=r^i \circ\phi(p) = x^i(p) =x^i$ where the last $x^i$ is a coordinate. We then have
$$\frac{\partial r^i}{\partial r^j}(\phi (p))=\frac{\partial x^i}{\partial r^j}.$$
Why does $\frac{\partial x^i}{\partial r^j} =\delta ^i_j$?
 A: You observe a certain ambiguity in Tu's notation when you write

It seems that the author is using $x^i$ and $r^i $ as functions in the numerator and as coordinates in the denominator.

This observation is correct; see also Section 6.6. Tu's Introduction to manifolds, definition of partial derivative.
Let us recall what the expression $\dfrac{\partial f}{\partial x^i} (p)$ means. Given a chart $\phi : U \to V$, where $p  \in U$ and $f : U \to \mathbb R$ is smooth, Tu defines
$$\dfrac{\partial f}{\partial x^i} (p) = \dfrac{\partial(f \circ \phi^{-1})}{\partial r^i} (\phi(p)) . \tag{1}$$
Here the symbols $r^i$ are used to denote the real coordinates of a point $\mathbf r \in V$. So the expression on the RHS of $(1)$ is an ordinary partial derivative in the sense of multivariable calclus. In textbooks on multivariable calculus these coordinates are usually denoted by $x_i$, but to avoid confusion Tu distinguishes ordinary real coordinates notationally from "generalized coordinates on a manifold".
By an abuse of notation Tu also writes $r^i$ for the $i$-th coordinate function $r_i : V \to \mathbb R$. More pedantically one should use something else, e.g. $\bar r^i : V \to \mathbb R, \bar r^i(r^1,\ldots,r^n) = r^i$. But if we are aware of this, it should not be a problem. As you said, if $r^i$ occurs in the numerator, then it has to be regarded as a function, and if it occurs in the denominator, as a coordinate variable. Now we get in fact
$$\dfrac{\partial r^i}{\partial r^j} (\mathbf r) = \delta^i_j$$
for all $\mathbf r \in V$.
Your argument

Using this definition, the numerator $r^i(\phi(p))=r^i \circ\phi(p) = x^i(p) =x^i$ where the last $x^i$ is a coordinate. We then have
$$\frac{\partial r^i}{\partial r^j}(\phi (p))=\frac{\partial x^i}{\partial r^j}.$$

is invalid. The statement "$r^i(\phi(p))=r^i \circ\phi(p) = x^i(p) =x^i$ where the last $x^i$ is a coordinate" is correct, but it does not allow to conclude
$$\dfrac{\partial r^i}{\partial r^j}(\phi (p))=\dfrac{\partial x^i}{\partial r^j}. \tag{2}$$
On the LHS we have the standard partial derivative $\dfrac{\partial r^i}{\partial r^j}$ at the point $\phi(p) \in V$, on the RHS we have an undefined expression. In fact, $x^i$ is a function on $U$ which does not allow to form $\dfrac{\partial x^i}{\partial r^j}$.
Let us finally discuss the local coordinates $x^i$. Tu defines local coordinate functions by
$$x^i = r^i \circ \phi : U \to \mathbb R .$$
It is clear that these depend on the chart $\phi$, thus it would perhaps be better to denote them by $x_\phi^i$. But this is unusual and would be some sort of pedantry. Again Tu does not notationally distinguish between the coordinate functions $x^i$ and the coordinates $x^i(p)$ of a point $p \in U$. Anyway, we have to be aware that the expression $\dfrac{\partial f}{\partial x^i} (p)$ in $(1)$ does not have an instrinsic meaning independent from the chart $\phi$; we need $(1)$ as a definition reducing it to something well-known from multivariable calculus. In particular
$$\dfrac{\partial x^i}{\partial x^j}$$
is a just a symbolic expression which needs an adequate interpretation via charts.
