Q1) Let $X$ a r.v. on a probability space $(\Omega ,\mathcal F,P)$ and $\mathcal G\subset \mathcal F$ a sub $\sigma -$algebra. What does $$\mathbb E[X\mid \mathcal G]\ \ ?$$ I understand what is $\mathbb E[X\mid Y]$ when $Y$ is an r.v., but not when it's a $\sigma -$algebra.
Q2) Let $(M_t)$ a stochastic process adapted to the filtration $\{\mathcal F_t\}_t$. I know that $(M_t)$ is a martingale if $\mathbb E[|M_t|]<\infty $ and $$\mathbb E[M_t\mid \mathcal F_s]=M_s,\quad s\leq t.$$ Could you explain me what does mean the last condition ? I would interpret is that when $t>s$, it's enough to know $M_s$ to now $M_t$ given $\mathcal F_s$, but I have no intuition neither any idea of what it can mean.
(Q1) I think it's helpful to think of it as an integral. For outcome $\omega\in\Omega$ we have that $\mathbb E [X \mid \mathcal G](\omega)=\frac{1}{P(A)}\int_A X(x) P(dx)$, where $A$ is the "smallest" set in $\mathcal G$ that contains $\omega$ (as long as $P(A)\neq0$). If $X$ is constant on $A$, then $\mathbb E [X \mid \mathcal G](\omega)=X(\omega).$
Another way to think of $\mathcal G$ is representing how accurately we can "resolve" the probability space. If we can distinguish between $X(\omega_1)$ and $X(\omega_2)$ for any two outcomes, then we have perfect resolution as far as $X$ is concerned. However, if $X(\omega_1)\neq X(\omega_2)$ but $\mathcal G$ doesn't contain any sets that have $\omega_1$ but not $\omega_2$ and vice versa (i.e. $\omega_1\in A$ implies $\omega_2\in A$ and vice versa), then if $\mathcal G$ defines our "resolution" we cannot distinguish between $\omega_1$ and $\omega_2$. Hence even though $X(\omega_1)\neq X(\omega_2)$ we have that $\mathbb E [X \mid \mathcal G](\omega_1)=\mathbb E [X \mid \mathcal G](\omega_2)$.
I find the picture on the Wikipedia article very helpful. Where you can see that $\mathbb E [X \mid \mathcal G]$ is just $X$ but averaged according to how well $\mathcal G$ resolves the probability space.
There are lots of other good answers to similar questions on MathSE as well if you search around.
(Q2) Regarding the Martingale property, consider a simple example: a gambler repeatedly flips a fair coin and gains $1$ dollar on heads and loses $1$ dollar on tails. Let $X_n$ be the gambler's balance after the $n^{th}$ coin flip. We can see that the expected winnings on the next flip is just the current balance of the gambler, but lets look at it regarding filtrations.
Let $X_0=0$ be the gamblers initial balance, then $X_1$ can be either $-1$ or $1$ hence $\sigma(X_1)=\left\{\emptyset, \{X_1=-1\}, \{X_1=1\}, \Omega\right\}.$ Note that $\{X=i\}$ means $\{\omega\in\Omega \mid X(\omega)=i\}.$
In general, we can show that $\mathbb E[X_{n+m}\mid\sigma(X_1,\ldots,X_n)]=X_n.$ This means that the distribution of the random variable $\mathbb E[X_{n+m}\mid\sigma(X_1,\ldots,X_n)]$ is the same as the distribution of $X_n.$
(Edit: actually it means a bit more than that: $\mathbb E[X_{n+m}\mid\sigma(X_1,\ldots,X_n)]$ and $X_n$ are the same random variable. This is a subtle, but important point. If I write $X=Y$ and perform the random experiment that gives $X=5$, then you know precisely that $Y=5$ as well, since they are the same random variable. But merely having the same distribution only gives $P(X=5)=P(Y=5)$, but they could be independent or have some arbitrary dependence structure, i.e. an arbitrary joint distribution---as long as it has the appropriate marginals.)
As a simple example, consider $\mathbb E[X_2\mid\sigma(X_1)].$ If $X_1(\omega)=-1$ then $E[X_2\mid\sigma(X_1)](\omega)=-1$ since $X_2$ has equal chances of being $0$ and $-2$, and if $X_1(\omega)=1$ then $E[X_2\mid\sigma(X_1)](\omega)=1$ since $X_2$ has equal chances of being $0$ and $2$. Thus $E[X_2\mid\sigma(X_1)]$ behaves exactly like $X_1.$ Of course, this is a very simple example. One could go on to show that $E[X_3\mid\sigma(X_1)]=X_1,$ $E[X_3\mid\sigma(X_1,X_2)]=X_2,$ etc. They aren't very difficult calculations and are quite instructive.