Isometry condition in components form From the book of Nakahara "Geometry, Topology and Physics":

Let $(M,g)$ be a (pseudo-) Riemannian manifold. A diffeomorfism
  $f:\mathcal{M}\to \mathcal{M}$ is an isometry if it preserves the
  metric: $$ f^{*}g_{f(p)}=g_{p} $$
In components this condition becomes: $$ \frac{\partial
 y^{\alpha}}{\partial x^{\mu}}\frac{\partial y^{\beta}}{\partial
 x^{\nu}}g_{\alpha\beta}(f(p))=g_{\mu\nu}(p) $$ where $x$ and $y$ are
  the coordinates of $p$ and $f(p)$, respectively.

I don't understand how he got the component form of the isometry condition $f^{*}g_{f(p)}=g_{p}$.
It reminds me the transformation law of a tensor (in this case the metric $g$):
$$ \frac{\partial
 x'^{\alpha}}{\partial x^{\mu}}\frac{\partial x'^{\beta}}{\partial
 x^{\nu}}g'_{\alpha\beta}(x')=g_{\mu\nu}(x) $$
when we change coordinates of the point $p\in M$ from $x$ to $x'$. But here the coordinates $x$  and $x'$ are referring to the same point $p$ the manifold $M$. In the isometry condition we have two different points on $M$, $p$ with coordinates $x$ and $f(p)$ with coordinates $y$. How are these two related? Is a diffeomorfism ("active" transformation - "moving" points on the manifold) the same thing as a coordinate transformation of the same point ("passive" transformation).
 A: Recall that the pullback is given by
$$(f^\ast g)_{p}(u, v) = g_{f(p)}(df_p(u), df_p(v))$$
for $u, v \in T_p M$. So if we have local coordinates $(x^\alpha)$, the components of $f^\ast g$ at $p \in M$ are
\begin{align}
(f^\ast g)_{\alpha\beta}(p) & = f^\ast g_{p}\left( \frac{\partial}{\partial x^\alpha}, \frac{\partial}{\partial x^\beta}\right) \\
 & = g_{f(p)}\left( df_p\left(\frac{\partial}{\partial x^\alpha}\right), df_p\left(\frac{\partial}{\partial x^\beta}\right)\right) \\
 & = g_{f(p)}\left( \frac{\partial f^\mu}{\partial x^\alpha}\frac{\partial}{\partial x^\mu}, \frac{\partial f^\nu}{\partial x^\beta}\frac{\partial}{\partial x^\nu}\right) \\
 & = \frac{\partial f^\mu}{\partial x^\alpha}\frac{\partial f^\nu}{\partial x^\beta}g_{f(p)}\left( \frac{\partial}{\partial x^\mu}, \frac{\partial}{\partial x^\nu}\right) \\
 & = \frac{\partial f^\mu}{\partial x^\alpha}\frac{\partial f^\nu}{\partial x^\beta}g_{\mu\nu}(f(p)).
\end{align}
Note that we only use the coordinate system $(x^\alpha)$ at $p$ here. To get the coordinates at $f(p)$, we just pull back the coordinates at $p$, i.e. we use the coordinates $(x^\alpha \circ f^{-1})$ at $f(p)$.
