Does the pullback of the metric tensor have the form $\phi^* g(X,Y)=g(\phi X,\phi Y)$? I have seen in many textbooks that the pullback of an arbitrary tensor field of type (r,s) under the diffeomorphism $\phi:M \rightarrow N$ is defined as
$\phi^* T(\eta_1,\dots, \eta_r, X_1, \dots, X_s) = T( (\phi^{-1})^*(\eta_1), \dots, (\phi^{-1})^*(\eta_r), \phi_* X_1, \dots, \phi_* X_s)$
where $\eta_i \in T_p^*(M)$ is a covector and $X_j \in T_p(M)$ is a vector. So, in the case of the metric tensor this would reduce to the following:
$\phi^*g(X,Y) = g(\phi_*X, \phi_*Y)$
where $X,Y$ are vectors.
Now, at the same time, Wikipedia suggests that we could find the pullback of such a tensor as
$\phi^*g(X,Y) = g(\phi X, \phi Y)$
and my question is how do they get $\phi_*$ to become $\phi$?
 A: If $\phi:V \to W$ is a linear map, then for any $p\in V$, the tangent mapping/pushforward mapping $T\phi_p$ or $d\phi_p$ or $\phi_{*,p}$ (however you want to use the notation) is a linear map $T_pV \to T_{\phi(p)}W$. But for a vector space, the tangent space can be canonically identified with itself: $T_pV \cong V$ and $T_{\phi(p)}W\cong W$. Because of this, you can "think of" the tangent mapping as a map $V \to W$. This is simply the derivative of a linear transformation $\phi:V \to W$ at the point $p \in V$. But a linear transformation is its own derivative.

If you want a more precise formulation of what I said above, here it is: on any (say finite-dimensional) vector space $V$, and any $p \in V$, there is a canonical isomorphism $\xi_{V,p}:T_pV \to V$. Note that the exact construction of this isomorphism will depend on which definition of tangent space you're using, but in any case, it is a good idea to prove this yourself. Similarly we have an isomorphism $\xi_{W,\phi(p)}:T_{\phi(p)}W \to W$. If you unwind the definitions of everything, you'll see that the following diagram commutes:
$\require{AMScd}$
\begin{CD}
T_pV @>{\phi_{*p}}>> T_{\phi(p)}W \\
@V{\xi_{V,p}}VV @VV{\xi_{W,\phi(p)}}V \\
V @>>{D\phi_p = \phi}> W
\end{CD}
In other words, $\phi = \xi_{W,\phi(p)} \circ \phi_{*,p} \circ (\xi_{V,p})^{-1}$, or said differently once again, up to isomorphisms, for each $p \in V$, we have $\phi_{*,p} = \phi$. But all of this is only because $\phi$ is a linear transformation.

But in the general case, if you have smooth manifolds $M,N$, and you have a metric tensor $g$ on $N$ and a diffeomorphism $\phi:M \to N$, there is no reason to even expect that $M,N$ have vector space structures, so it doesn't even make sense to talk about $\phi$ being linear. This is why we have to use the push-forward map, and there is no sense in which we can "identify" the push-forward with the original map itself.
See this for a more general perspective of everything I mentioned here (with slightly different notation).

Edit: In response to comment.
The author DOES NOT say $(s^{-1})^*g(x,y) = g(s^{-1}(x), s^{-1}(y))$. He says
\begin{align}
d_{B^n}(x,y) &= [(s^{-1})^*d_{\mathcal{H}^n}](x,y) = d_{\mathcal{H}^n}(s^{-1}(x), s^{-1}(y))
\end{align}
These are completely different statements. Note that if you have two (let's for simplicity say simply connected) Riemannian manifolds $(M,g)$ and $(N,h)$. Then, the metric tensors $g$ and $h$ give rise to distance functions $d_g$ and $d_h$ respectively (in the article, the author refers to these as $d_{B^n}$ and $d_{\mathcal{H}^n}$). Now, suppose we have a diffeomorphism $\phi:M \to N$. Then, we can consider the following objects:

*

*first is the pullback tensor field $(\phi^{-1})^*g$ on $N$ (as defined above).

*second is the pull-back distance function $(\phi^{-1})^*d_g$ on $N$, which is DEFINED as
\begin{align}
[(\phi^{-1})^*d_g](x,y) &= d_g\left( \phi^{-1}(x), \phi^{-1}(y)\right) \qquad \text{for all $x,y \in N$}
\end{align}
Note that although we are using the same notation $(\phi^{-1})^*$, and calling both of them "pullbacks", these are completely different things. The first is a pullback of tensor field, while the second is a pull-back of a distance function. The word "pull-back" should be thought of literally as the name suggests: you have a certain object defined on one space (eg. a tensor field or distance function), and you have a invertible map between two spaces. Then, you can use this map to "transport" this object to the new space.
Now, here is a theorem which you should try to prove (it is really just an exercise in unwinding all the definitions).

Theorem.
Let $(M,g), (N,h)$, $\phi:M \to N$, and $d_g,d_h$ all have the same definitions as above. If $h = (\phi^{-1})^*g$ then $d_h = (\phi^{-1})^*d_g$.

What this says is that if your metric tensors are related to each other by a pullback, then so are the associated distance functions. Note that this is precisely what the author is saying in the first sentence of his proof:

"Since the stereographic projection $s: \mathcal{H}^+ \to B^n$ is a Riemannian isometry, it is also a metric isometry for the induced distances."

