"In classical mechanics, momentum is defined as the quantity which is conserved under global spatial translations or, alternatively, as the generator of spatial translations." (G.Parisi, Quantum mechanics)
From this definition (the generator of spatial translation), how can I obtain the analytical expression of the momentum (i.d. $p=mv$)?
If we have some coordinates $q_i$ and some momenta $p_i$, then a generator of a transformation is defined as a function $g(q_i, p_i)$. By definition, this generates the transformation
$$q_i \to q_i + \epsilon \frac{\partial g}{\partial p_i}$$
$$p_i \to p_i + \epsilon \frac{\partial g}{\partial q_i}$$
So if we want the generator of translations, we want
$$q_i \to q_i + \epsilon$$
where $q_i = x$, some particular rectangular coordinate, and also
$$p_x \to p_x$$
(since we want to generate only translation without changing momenta). These imply $g = p_x + c$. Setting $c=0$, we see that x-momentum is the generator of translations in the x-direction.
You can't really show $p_x = m\dot{x}$ unless you make some assumptions about the Hamiltonian. If we assume $H(x,p_x) = \frac{1}{2m}p_x^2 + V(x)$, then Hamilton's equations give
$$\frac{\partial H}{\partial p_x} = \dot{x} = \frac{p}{m}$$
as you requested.
All the above will only make sense if you've studied analytical mechanics from a source that happened to leave out generators. If not, you'll probably want to review the Hamiltonian formulation of mechanics. Chapter 2 of Shankar's Principles of Quantum Mechanics is an overview of analytical mechanics specifically aimed at getting you ready for quantum mechanics and includes a discussion of generators.
