How Heisenberg's uncertainty principle reviews the classical definition of a point?

I'm not entirely sure I understand what you're trying to ask, but I'll make a stab. From a modern perspective, the uncertainty principle comes in two steps: (i) We allow, or even demand, noncommutativity in our algebras of observables. (ii) Fix a state $\phi$, normalized so that $\langle \phi | \phi \rangle = 1$. If $f$ is any observable, the expectation value of $f$ in state $\phi$ is $\langle f \rangle = \langle \phi | f | \phi \rangle$. Note that we allow everything to be complex-valued; denote complex (i.e. Hermitian) conjugation by $\Box \mapsto \Box^\ast$. The uncertainty of $f$ in state $\phi$ is $\Delta f = \sqrt{ \langle f^\ast f \rangle - \langle f \rangle \langle f^\ast \rangle} = \sqrt{ \bigl\langle ( f - \langle f\rangle)^\ast (f - \langle f \rangle) \bigr\rangle }$. Let $f,g$ be any observables. The Cauchy-Schwartz inequality then says that: $$ \begin{align} (\Delta f)^2(\Delta g)^2 & = \bigl\langle (f - \langle f \rangle)\phi \big| (f - \langle f \rangle)\phi \bigr\rangle \bigl\langle (g - \langle g \rangle)\phi \big| (g - \langle g \rangle)\phi \bigr\rangle \\\ & \geq \bigl\langle (f - \langle f \rangle)\phi \big| (g - \langle g \rangle)\phi \bigr\rangle \bigl\langle (g - \langle g \rangle)\phi \big| (f - \langle f \rangle)\phi \bigr\rangle \\\ & = \bigl\langle \phi \big| (f - \langle f \rangle)^\ast (g - \langle g \rangle) \big| \phi \rangle \bigl\langle \phi \big| (f - \langle f \rangle)^\ast (g - \langle g \rangle) \big| \phi \rangle^\ast \\\ (\Delta f) (\Delta g) & \geq \bigl| \bigl\langle (f - \langle f\rangle)^\ast (g - \langle g \rangle) \bigr\rangle \bigr| \end{align} $$ Note that $\Delta f = \Delta(f^\ast)$. Set $f' = f - \langle f \rangle$ and $g' = g - \langle g \rangle$. Then, switching $f \leadsto f^\ast$ in the above calculation, we've shown that $(\Delta f)(\Delta g) \geq | \langle f'g'\rangle|$, and also $\geq | -\langle g'f'\rangle|$, adding a minus sign just for fun. But then: $$ \begin{align} (\Delta f)(\Delta g) & \geq \frac12 \left( \bigl| \langle f'g'\rangle\bigr| + \bigl| -\langle g'f'\rangle\bigr|\right) \\\ & \geq \frac12 \left| \langle f'g'\rangle - \langle g'f'\rangle \right| \\\ & = \frac12 \left| \langle [f',g']\rangle \right| = \frac12 \left| \langle [f,g]\rangle \right| \end{align} $$ by the triangle inequality. (Note that you can also get a lower bound of $\frac12 \bigl| \langle f'g' + g'f' \rangle\bigr|$, and in fact you can show the following sharpening by Schrodinger: $(\Delta f)(\Delta g) \geq \sqrt{ \left( \frac12 \bigl| \langle [f,g]\rangle \bigr| \right)^2 + \left( \frac12 \bigl| \langle f'g' + g'f' \rangle\bigr| \right)^2 }$.)

Anyway, the particular example is whenever $f,g$ are in canonical commutation, i.e. $[f,g]$ is a nonzero constant (usually fixed at Planck's constant, up to some $2\pi i$s). Then $\langle [f,g]\rangle$ is the expectation value of a constant, and so is independent of the state $\phi$. This proves that as soon as you have such observables in your algebra, then there cannot be states for which all uncertainties vanish. This is in contrast with the commutative situation: a "point" might be defined as a state in which all uncertainties vanish, and in a (well-behaved) commutative algebra every state is a convex combination of points.

So I guess the answer to your first question is that Uncertainty requires that you abandon the notion of "point" in noncommutative spaces, in favor of a more spread-out notion of "state". In commutative land, it suffices to consider the "points", but in noncommutative land there generically are no points at all.

Why deformation always starts with a Poisson manifold? if it is to deform the phase space of hamiltonian mechanics it suffices to consider symplectic manifolds!

This question, I think, starts from a mistaken impression of physical systems. At first, students study physical systems that are symplectic — indeed, one begins only with cotangent bundles, being the ones that describe physical systems with a good notion of "configuration space" or "position space". Note that none of this complicated machinery of deformation quantization is necessary there: if you have a configuration space $N$ with a distinguished volume measure, then you can immediately write down the $L^2$ functions on $N$ as your Hilbert space, and the (bounded operators within the appropriate completion of) the differential operators as your algebra. The algebra of differential operators is naturally filtered, and its associated graded is canonically isomorphic to the algebra of functions on the cotangent bundle; so the Rees algebra construction without effort introduces Planck's constant into the game and lets you take classical limits, etc. (If $N$ does not have a measure, then you still have the algebra of differential operators, which acts on the vector space of functions, but that vector space does not have a Hilbert space structure; you also have the Hilbert space of $L^2$ half-densities on $N$, and its algebra of bounded operators, but you do not have a canonical "classical limit" of this algebra.)

But there are many more complicated physical systems, and they are not all symplectic. Indeed, Dirac recognized early on that constrained mechanical systems can be well-described by possibly-non-symplectic Poisson manifolds (maybe with singularities). He talks about "first class constraints" and "second class constraints", and I don't remember which is which, but one of them is constraints that preserve symplecticity and the other do not, if memory serves. There's probably a discussion of this in the book by Marsden and Weinstein, although I don't have it in front of me so I can't check.

Why the deformation of the algebra of functions $C^\infty(M)$ of a Poisson manifold is a way of quantization? in this case what is the Hilbert space, in which the observable $f$ are replaced by a bounded operators $\hat f$?

Ah, this I think is the most interesting of your questions. The answer is: there isn't one. The GNS construction assures that for noncommutative algebras with good analytic control, there always exists a Hilbert space on which the algebra is faithfully represented, although it's not (in most write-ups) at all canonical. (Deformation quantization does not produce algebras with sufficient analytic control, but I will sweep that issue under the rug.) This is well-illustrated by the example of cotangent bundles I discussed above. Let $N$ be a manifold and $M = \mathrm T^\ast N$ its cotangent bundle. A very good deformation quantization of $C^\infty(M)$ is the Rees algebra of differential operators on $N$: it is generated by smooth functions on $N$ and $\hbar\times$ vector fields on $N$, and then you may have to do some analytic work if you really want to be a deformation of $C^\infty$ and not just of the part that's polynomial in the $\mathrm T^\ast$-direction. This deformation quantization always exists, and always acts on $C^\infty(N)$. But $C^\infty(N)$ is not a Hilbert space without more choices: namely, you must choose a measure on $N$ to be able to define $\langle f,g\rangle = \int_N f^\ast g$. A choice of measure essentially amounts to a choice of state for the GNS construction applied to the algebra of differential operators.

Note furthermore that a priori, there also is not a canonical map $C^\infty(M) \to $ differential operators, at least not without choices. Kontsevich's result that you alluded to above picks out such a map, but does require a choice (that said, I get myself confused about how much of a choice it requires).

The point is that for many questions, just having a noncommutative algebra with some parameter $\hbar$, whose $\hbar \to 0$ limit is your commutative algebra (and that's maybe "flat" over the space of values of $\hbar$) is enough for many applications. The early quantum mechanists (and still all textbooks) were simply asking too much to have a Hilbert space and a quantization map and all that — at least, that's my opinion about it. The buzzwords in physics books are "Schrodinger picture" and "Heisenberg picture": in "Schrodinger picture" the Hilbert space is actually a physical, existing thing, whereas in "Heisenberg picture" basically all you have is the algebra of observables. The Heisenberg picture is much more philosophically reasonable: as scientists, we make observations and measurements; we don't directly manipulate vectors in a Hilbert space. If possible, it's best to set up a mathematical framework that's as minimal and faithful-to-what-people-actually-do (not what they say they do) as possible.

How Heisenberg's uncertainty principle reviews the classical definition of a point?

I'm not entirely sure I understand what you're trying to ask, but I'll make a stab. From a modern perspective, the uncertainty principle comes in two steps: (i) We allow, or even demand, noncommutativity in our algebras of observables. (ii) Fix a state $\phi$, normalized so that $\langle \phi | \phi \rangle = 1$. If $f$ is any observable, the expectation value of $f$ in state $\phi$ is $\langle f \rangle = \langle \phi | f | \phi \rangle$. Note that we allow everything to be complex-valued; denote complex (i.e. Hermitian) conjugation by $\Box \mapsto \Box^\ast$. The uncertainty of $f$ in state $\phi$ is $\Delta f = \sqrt{ \langle f^\ast f \rangle - \langle f \rangle \langle f^\ast \rangle} = \sqrt{ \bigl\langle ( f - \langle f\rangle)^\ast (f - \langle f \rangle) \bigr\rangle }$. Let $f,g$ be any observables. The Cauchy-Schwartz inequality then says that: $$ \begin{align} (\Delta f)^2(\Delta g)^2 & = \bigl\langle (f - \langle f \rangle)\phi \big| (f - \langle f \rangle)\phi \bigr\rangle \bigl\langle (g - \langle g \rangle)\phi \big| (g - \langle g \rangle)\phi \bigr\rangle \\\ & \geq \bigl\langle (f - \langle f \rangle)\phi \big| (g - \langle g \rangle)\phi \bigr\rangle \bigl\langle (g - \langle g \rangle)\phi \big| (f - \langle f \rangle)\phi \bigr\rangle \\\ & = \bigl\langle \phi \big| (f - \langle f \rangle)^\ast (g - \langle g \rangle) \big| \phi \rangle \bigl\langle \phi \big| (f - \langle f \rangle)^\ast (g - \langle g \rangle) \big| \phi \rangle^\ast \\\ (\Delta f) (\Delta g) & \geq \bigl| \bigl\langle (f - \langle f\rangle)^\ast (g - \langle g \rangle) \bigr\rangle \bigr| \end{align} $$ Note that $\Delta f = \Delta(f^\ast)$. Set $f' = f - \langle f \rangle$ and $g' = g - \langle g \rangle$. Then, switching $f \leadsto f^\ast$ in the above calculation, we've shown that $(\Delta f)(\Delta g) \geq | \langle f'g'\rangle|$, and also $\geq | -\langle g'f'\rangle|$, adding a minus sign just for fun. But then: $$ \begin{align} (\Delta f)(\Delta g) & \geq \frac12 \left( \bigl| \langle f'g'\rangle\bigr| + \bigl| -\langle g'f'\rangle\bigr|\right) \\\ & \geq \frac12 \left| \langle f'g'\rangle - \langle g'f'\rangle \right| \\\ & = \frac12 \left| \langle [f',g']\rangle \right| = \frac12 \left| \langle [f,g]\rangle \right| \end{align} $$ by the triangle inequality. (Note that you can also get a lower bound of $\frac12 \bigl| \langle f'g' + g'f' \rangle\bigr|$, and thus by the triangle inequality you have the following sharpening by Schrodinger: $(\Delta f)(\Delta g) \geq \sqrt{ \left( \frac12 \bigl| \langle [f,g]\rangle \bigr| \right)^2 + \left( \frac12 \bigl| \langle f'g' + g'f' \rangle\bigr| \right)^2 }$. Note also that many accounts give the much weaker but still useful estimate that $(\Delta f)(\Delta g) \geq \frac12 \operatorname{Im}\langle [f,g]\rangle$.)

Anyway, the particular example is whenever $f,g$ are in canonical commutation, i.e. $[f,g]$ is a nonzero constant (usually fixed at Planck's constant, up to some $2\pi i$s). Then $\langle [f,g]\rangle$ is the expectation value of a constant, and so is independent of the state $\phi$. This proves that as soon as you have such observables in your algebra, then there cannot be states for which all uncertainties vanish. This is in contrast with the commutative situation: a "point" might be defined as a state in which all uncertainties vanish, and in a (well-behaved) commutative algebra every state is a convex combination of points.

So I guess the answer to your first question is that Uncertainty requires that you abandon the notion of "point" in noncommutative spaces, in favor of a more spread-out notion of "state". In commutative land, it suffices to consider the "points", but in noncommutative land there generically are no points at all.

Why deformation always starts with a Poisson manifold? if it is to deform the phase space of hamiltonian mechanics it suffices to consider symplectic manifolds!

This question, I think, starts from a mistaken impression of physical systems. At first, students study physical systems that are symplectic — indeed, one begins only with cotangent bundles, being the ones that describe physical systems with a good notion of "configuration space" or "position space". Note that none of this complicated machinery of deformation quantization is necessary there: if you have a configuration space $N$ with a distinguished volume measure, then you can immediately write down the $L^2$ functions on $N$ as your Hilbert space, and the (bounded operators within the appropriate completion of) the differential operators as your algebra. The algebra of differential operators is naturally filtered, and its associated graded is canonically isomorphic to the algebra of functions on the cotangent bundle; so the Rees algebra construction without effort introduces Planck's constant into the game and lets you take classical limits, etc. (If $N$ does not have a measure, then you still have the algebra of differential operators, which acts on the vector space of functions, but that vector space does not have a Hilbert space structure; you also have the Hilbert space of $L^2$ half-densities on $N$, and its algebra of bounded operators, but you do not have a canonical "classical limit" of this algebra.)

But there are many more complicated physical systems, and they are not all symplectic. Indeed, Dirac recognized early on that constrained mechanical systems can be well-described by possibly-non-symplectic Poisson manifolds (maybe with singularities). He talks about "first class constraints" and "second class constraints", and I don't remember which is which, but one of them is constraints that preserve symplecticity and the other do not, if memory serves. There's probably a discussion of this in the book by Marsden and Weinstein, although I don't have it in front of me so I can't check.

Why the deformation of the algebra of functions $C^\infty(M)$ of a Poisson manifold is a way of quantization? in this case what is the Hilbert space, in which the observable $f$ are replaced by a bounded operators $\hat f$?

Ah, this I think is the most interesting of your questions. The answer is: there isn't one. The GNS construction assures that for noncommutative algebras with good analytic control, there always exists a Hilbert space on which the algebra is faithfully represented, although it's not (in most write-ups) at all canonical. (Deformation quantization does not produce algebras with sufficient analytic control, but I will sweep that issue under the rug.) This is well-illustrated by the example of cotangent bundles I discussed above. Let $N$ be a manifold and $M = \mathrm T^\ast N$ its cotangent bundle. A very good deformation quantization of $C^\infty(M)$ is the Rees algebra of differential operators on $N$: it is generated by smooth functions on $N$ and $\hbar\times$ vector fields on $N$, and then you may have to do some analytic work if you really want to be a deformation of $C^\infty$ and not just of the part that's polynomial in the $\mathrm T^\ast$-direction. This deformation quantization always exists, and always acts on $C^\infty(N)$. But $C^\infty(N)$ is not a Hilbert space without more choices: namely, you must choose a measure on $N$ to be able to define $\langle f,g\rangle = \int_N f^\ast g$. A choice of measure essentially amounts to a choice of state for the GNS construction applied to the algebra of differential operators.

Note furthermore that a priori, there also is not a canonical map $C^\infty(M) \to $ differential operators, at least not without choices. Kontsevich's result that you alluded to above picks out such a map, but does require a choice (that said, I get myself confused about how much of a choice it requires).

The point is that for many questions, just having a noncommutative algebra with some parameter $\hbar$, whose $\hbar \to 0$ limit is your commutative algebra (and that's maybe "flat" over the space of values of $\hbar$) is enough for many applications. The early quantum mechanists (and still all textbooks) were simply asking too much to have a Hilbert space and a quantization map and all that — at least, that's my opinion about it. The buzzwords in physics books are "Schrodinger picture" and "Heisenberg picture": in "Schrodinger picture" the Hilbert space is actually a physical, existing thing, whereas in "Heisenberg picture" basically all you have is the algebra of observables. The Heisenberg picture is much more philosophically reasonable: as scientists, we make observations and measurements; we don't directly manipulate vectors in a Hilbert space. If possible, it's best to set up a mathematical framework that's as minimal and faithful-to-what-people-actually-do (not what they say they do) as possible.