Suppose $M\subseteq X$ and we consider the probability spaces $(M,\mathcal{B}(X)_{|M},\mu)$, where $\mathcal{B}(X)$ is the Borel-$\sigma$-algebra of $X$, and $(X,\mathcal{B}(X),\nu)$, where $\nu:=\iota_*(\mu)$ is the pushforward measure of $\mu$ under the embedding $\iota: M\to X$ with $\iota^{-1}(B)=M\cap B$ for $B\in\mathcal{B}(X)$.
Now, let $\alpha=\{\alpha_1,\ldots,\alpha_k\}$ be any finite, measurable partition of $M$ and $\beta=\{\beta_1,\ldots,\beta_m\}$ any finite measurable partition of $X$.
Isn't it true that, up to sets of measure $\nu=0$, $$ \alpha = \beta $$
and in particular that for a continuous map $f\colon X\to X$ the partions $$ \bigvee_{i=0}^nf^{-i}\beta:=\left\{\bigcap_{i=0}^{n}f^{-i}\beta_{j_i}: j_i\in \{1,\ldots,m\}\right\} $$ and $$ \bigvee_{i=0}^{n}(f_{|M})^{-i}\alpha:=\left\{\bigcap_{i=0}^{n}(f_{|M})^{-1}\alpha_{j_i}: j_i\in\{1,\ldots,k\}\right\} $$ coincide up to sets which have measure $\nu=0$?
(Here with $f_{|M}$ I mean the restriction of $f$ onto $M$ and I suppose that $f(M)\subseteq M$, i.e. $f$ is invariant on $M$.)
I am asking this because if this is true then for the measure-theoretic entropies of $f$ on $(X,\mathcal{B}(X),\nu)$ and of $f_{|M}$ on $(M,\mathcal{B}(X)_{|M},\mu)$, this would imply that these two entropies coincide, wouldn't it?
You have not hypothesized any relation whatsoever between $\alpha$ and $\beta$, so this is not true.
For example let $M=[0,1] \subset [0,2]=X$ and let $\mu$ be Lebesgue measure on $[0,1]$, so $\nu$ assigns measure zero to $(1,2]$. We could take $\alpha = \{[0,1/2),[1/2,1]\}$ and $\beta = \{[0,3/4],(3/4,2]\}$ for a counterexample.
It should nonetheless be true that those two measure-theoretic entropies of $f$ in your last paragraph do coincide, but not for the reasons you've given. I would suggest instead investigating comparisons between $\bigvee_{i=0}^nf^{-i}\beta$ and $\bigvee_{i=0}^{n}(f_{|M})^{-i}\alpha$ in two special situations: