I've got a definition, but says something strange, what does it mean? $M_1:=(M,d_1), \ \  M_2:=(M,d_2)$.  $d_1$ is equivalent to $d_2$ if the identity $x\rightarrow x$ of $M_1$ over $M_2$ is an homeomorphism
I'm not sure what it is talkin about when it says "identity" becasue the "$x\rightarrow x$" confuses me a bit, also, if it's an identity why does it point out 2 different sets with the word "over" (in bold)?

The word "over" in the quoted definition sounds strange; so let's forget about it.
A homeomorphism  $f: X \to Y$ is a bijective map between toplogical spaces whereby both $f$ and $f^{-1}$ are continuous. This is the same as a bijective map whereby both $f$ and $f^{-1}$ send open sets to open sets.
In the example at hand one and the same "ground set" $M$ carries two different metrics $d_1$ and $d_2$. Each of these metrics establishes on $M$ a corresponding topology that determines which sets $A\subset M$ are open, or which functions $g:\>M\to{\mathbb R}$ are continuous, etc. The two metrics are called equivalent if they induce the same toplogy on $M$. This means that the measured distances may be very different for the two metrics, but when it comes to openness of sets $A$ or continuity of functions on $M$ there is no difference.
This is the case iff any $d_1$-neighborhood $U^{d_1}_\epsilon(p):=\{x\in M\>|\> d_1(x,p)<\epsilon\}$ of any point $p\in M$ contains a $d_2$-neighborhood $U^{d_2}_{\epsilon'}(p)$ for a suitable $\epsilon'$, and vice versa.
Now a fancy way of expressing this condition is the following: The map $$\iota: \quad(M,d_1)\to(M,d_2),\qquad x\mapsto x$$
is a homeomorphism.
