About reference frame in Newton's second law? Classical physics models events occuring in the spacetime $\mathcal E\times \mathcal T$ where $\mathcal E$ is a dimension 3 euclidean point space and 
$\mathcal T$ is an interval of $(\mathbb R, <)$ (an ordered set).
An observer is a fictitious human being or sensor that can infinitely precisely describs events.
A reference frame is made of a frame $R=(O, \vec e_1, \vec e_2, \vec e_3)$,
an initial time $t_0$ and a coordinate system uniquely mapping a triplet $(x,y,z)$
to a generic point $M$ in $\mathcal E$.
Point $O$ is in $\mathcal E$ and is chosen as the origin and $(\vec e_1, \vec e_2, \vec e_3)$ is a basis of the euclidean vector space $\vec E$ attached to $\mathcal E$.
It seems to me that the first purpose of such a reference frame is to quantify
(I mean with values not with symbols) the location and the instant of any event. Without it a generic point $M$ still qualifies a unique location of $\mathcal E$ and a real $t$ an instant in $\mathcal T$ and the inner product a tool for geometry...
Since we can define arbitrarily a reference frame given a point and a basis
let's suppose that $O$ is the location of the observer. The observer moves with point $O$.
It is often said that Newton's first law postulates the existence of privilegied reference frames in which a body experiencing null resultant of forces $\vec F$ have a constant velocity, that is null acceleration $\vec a$, such frames are called inertial reference frames or galilean reference frames. Then Newton's second law as the simple form $\vec F=m \vec a$ only in inertial reference frames.
In mathematics, it is taught that a vector is intrinsic in the sense that
it's existence preceeds the one of basis and it is not dependent of the basis in which it is quantified.
So my question is why the notion of reference frame is usefull in the statement of Newton's second law?
It seems to me that only point $O$ is needed, we actually don't care about the basis. Point $O$ should be an "inertial point" the basis can rotate, vectors change in norm, angles between two vectors can change as long as they stay linearly independent because we still can quantity $\vec F$ in every such basis.
Do you have an opinion on that? Any book that consider vectors as intrinsic objects?
 A: You have a wrong idea of classical spacetime $V^4$. It is not the Cartesian product $\mathbb E^3 \times \mathbb R$. 
It is instead a fiber bundle $$T: V^4 \to \mathbb R$$
such that each fiber $\Sigma_t = T^{-1}(t)$, the absolute space at fixed time $t$,  is isomorphic to a three dimensional Euclidean space (*) $\mathbb E^3$. The basis of the bundle, the axis $\mathbb R$, is the range of the absolute time $T$ which is defined up to an additive constant.
The difference between this notion of spacetime as a bundle $T: V^4 \to \mathbb R$ and a trivial scalar product 
$\mathbb E^3 \times \mathbb R$ is fundamental: Here there is no canonical representation of $V^4$ as a Cartesian product $\mathbb E^3 \times \mathbb R$.
More precisely, every choice of a reference frame defines such a representation. 
A reference frame is nothing but a (smooth) surjective map $$\pi : V^4 \to \mathbb E^3$$ such that $\pi|_{\Sigma_t} : \Sigma_t \to \mathbb E^3$ is an isomorphism of Euclidean spaces (i.e., a surjective affine isometry), for every $t\in \mathbb R$. 
In this picture, $\mathbb E^3$ is viewed as the rest space of the reference frame.
This way the spacetime $V^4$ is identified with the Cartesian product $\mathbb R \times \mathbb E^3$ by means of
$$V^4 \ni p \mapsto (T(p), \pi(p)) \in \mathbb R \times \mathbb E^3$$
There are however infinitely many such identifications depending on the choice of the reference frame.
Consider two reference frames $\pi$ and $\pi'$ and fix Cartesian orthonormal coordinates in the respective rest spaces $\mathbb E^3$ and
$\mathbb E'^3$, and use the absolute time defined up to an additive constant as time coordinate.
Using the fact that  $\pi'|_{\Sigma_t}\circ (\pi|_{\Sigma_t})^{-1} : \mathbb E^3 \to \mathbb E'^3$ is  a surjective affine isometry,
 you easily see that the transformation of coordinates must be of the form
$$t'=t+c \:,\quad x'_i = \sum_{j=1}^3 R_{ij}(t) x_j + b_i(t) \tag{1}$$
where $R(t) \in O(3)$ and $b(t) \in \mathbb R$ for every $t$.
These are the most general transformation of coordinates between Cartesian coordinates at rest with different reference frames.
To define the velocity of a section $$\mathbb R \ni t \mapsto \gamma(t) \in \Sigma_t $$  you need a full reference frame as indicated not only a reference point.
Indeed, the velocity of $\gamma$ with respect to $\pi$ is computed as 
$$\vec{V}_\pi(t) := \frac{d}{dt}\pi(\gamma(t))$$
and, with this definition, it is a vector in the rest space $\mathbb E^3$ of $\pi$. However it can be seen as a vector in $\Sigma_t$ using the inverse of the isomprphism  $\pi|_{\Sigma_t} : \Sigma_t \to \mathbb E^3$. 
Exploiting this identification, velocities of the same section but referred to different reference frames can be compared in the absolute space $\Sigma_t$.
REMARK. It is not enough to fix a reference point, that is a section $\mathbb R \ni t \to O(t) \in \Sigma_t$ to define the velocity of another section $\mathbb R \ni t \to \gamma(t) \in \Sigma_t$. Your  idea is to take the limit $$\lim_{h \to 0} \frac{1}{h}\left[\left(\gamma(t+h) -O(t+h)\right) - \left(\gamma(t) -O(t)\right)\right]\:.$$
The point is that the difference $$\left(\gamma(t+h) -O(t+h)\right) - \left(\gamma(t) -O(t)\right)$$ does not make sense as the two vectors$\gamma(t+h) -O(t+h)$ and $\gamma(t) -O(t)$ belong to different vector spaces. In order to make sensible that difference is necessary to isometrically identify the spaces. This is exactly what the notion of reference frame does.
Inertial reference frames are defined as reference frames where every isolated body moves with constant velocity. It is easily proved that this constraint imposes a strong restriction to the form of the coordinate transformation (1) between inertial frames which therefore specialises to
$$t'=t+c \:,\quad x'_i = \sum_{j=1}^3 R_{ij} x_j + tv_i + b_i \tag{2}$$ 
that is a generic transformation of Galileo's group. It is nice to observe that, up to isomorphisms, there is only one affine structure in the classical spacetime such that Cartesian coordinates at rest with inertial reference frames together with absolute time as forth coordinate define affine coordinate systems of that structure. The sections of the spacetime which are right lines (geodesics) of that affine structure are all possible inertial evolutions of isolated matter points. To this respect classical physics and GR are not so different.
Statements as Newton's second law are formulated with this notion of reference frame (if one wants to be completely rigorous).

(*) An Euclidean  space $\mathbb E^n$ is an affine space whose $n$-dimensional space of vectors $T^n$ describing translations in $\mathbb E^n$ is equipped with a positive scalar product.
A: 
It seems to me that only point O is needed, we actually don't care about the basis.

We actually do care about the basis, very much so. In Newtonian mechanics, the displacement vector between one point and another in $\mathbb R^3$ is indeed frame independent. This displacement vector might well have different representations in different bases, but all such displacement vectors are substantively the same vector.
The same does not apply to the time derivatives of these vectors. The time derivative of a displacement vector is a frame-dependent quantity, depending on the linear and angular velocities of the observers. This also applies to the second time derivative of a displacement vector. Since Newton's second law is a statement about second time derivatives, that displacement vectors are substantively the same to all observers is irrelevant. 
