Be $\mathcal{C}$ a category, $M \in \mathcal{C}$ and $F$ a funtor of $\mathcal{C}$ to $\mathcal{C}_{Set}$. By Yoneda Lemma we have that Hom(Hom($M$,-),$F$)$=F(M)$. I understand in the proof that we have a biyection from Hom(Hom(M,-),F) to $F$ but I don`t understand the "$=$".
And I don´t understand that $M \simeq M' \Leftrightarrow \text{Hom(}M\text{,-)} \simeq \text{Hom(}M'\text{,-)}$.
The Yoneda Lemma says that there is a bijection $Hom(Hom(M,-),F)\cong F(M)$. The equal sign, if you have encountered the lemma written with it instead of the isomorphism sign, is an abuse of notation. It is very common to regard isomorphic objects as being the same. For instance, the groups $\{-1,1\}$ with the product and $\{0,1\}$ with the sum modulo $2$ are both isomorphic to the cyclic group of order two. They are not literally the same, but we can say that both groups are the cyclic group of order $2$. And the same happens a lot with any other equivalence relation (the actual realization of this idea is the quotient). Morever, the isomorphism (bijection in this case) is natural on both $M$ and $F$, so we can treat them as equal in many categorical circumstances.
Natural in $M$ means that a morphism $M\to M'$ induces a morphism $Hom(Hom(M,-),F)\to Hom(Hom(M',-),F)$, which is an isomorphism if $M\to M'$ is an isomorphism. Similarly the naturality in $F$.
As per the other question, there is a nice answer in this question. I think the non accepted answer might be more enlightening if you have problems understanding the Yoneda embedding.