Let $\Sigma$ be a compact submanifold of a Riemannian manifold $(M,g)$, let $h$ be the induced metric on $\Sigma$, and let $\Phi:M\to M$ be a diffeomorphism. $h_\Phi$ will denote the induced metric on the image $\Phi(\Sigma)$, pulled back to $\Sigma$ via $\Phi$.
What is the $h_\Phi$ really? I know $\Phi(\Sigma)$ is an embedded submanifold of $M$ with the property that $\Phi$ maps $\Sigma$ diffeomorphically onto $\Phi(\Sigma)$, which begs a question:do the Riemannian metrics on $\Phi(\Sigma)$ induced respectively by the inclusion $\iota|_{\Phi(\Sigma)}:\Phi(\Sigma)\hookrightarrow M$ and the restriction $\Phi|_\Sigma$ coincide? I supposed that $h_\Phi$ is the Riemannian metric induced by $\iota|_{\Phi(\Sigma)}$ and tried to show that $$\Phi|_\Sigma^*h_\Phi=h$$ but didn't find a way. The following is as far as I can go now: $$\Phi|_\Sigma^*h_\Phi=\Phi|_\Sigma^*(\iota_{\Phi(\Sigma)}^*g)=(\iota_{\Phi(\Sigma)}\circ\Phi|_\Sigma)^*g=???=\iota_\Sigma^*g=h$$ $\iota_\Sigma$ is the inclusion $\Sigma\hookrightarrow M$. Thanks for help.
Edit. I think the author was trying to say $$h_\Phi=(\Phi|_\Sigma^{-1})^*h.$$ That way, we would have $$h=\Phi|_\Sigma^*h_\Phi.$$ This is pretty much the scenario described in the quote.
$\Phi$ being a diffeomorphism $M\to M$, it induces a diffeomorphism from $\Sigma$ to $\Phi(\Sigma)$. Note that $\Phi(\Sigma)$ is a submanifold of $(M,g)$, so it has an induced metric $g|_{\Phi(\Sigma)}$. You can pull it back on $\Sigma$ by $\Phi$, yielding $h_{\Phi} = \Phi|_{\Sigma}^*( g|_{\Phi(\Sigma)})$. All these complicated notations to just talk about the following: $$ \forall p \in \Sigma, \forall u,v \in T_p\Sigma,\quad (h_{\Phi})(u,v) = g(d\Phi(p)u,d\Phi(p)v). $$