About global statement and local statement:
There is a lemma in variational calculus that was first stated by Jacob Bernoulli.
When Johann Bernoulli had presented the Brachistochrone problem to the mathematicians of the time Jacob Bernoulli was among the few who was able to find the solution independently. The treatment by Jacob Bernoulli is in the Acta Eruditorum, May 1697, pp. 211-217
Jacob opens his treatment with an observation concerning the fact that the curve that is sought is a minimum curve.
Lemma. Let ACEDB be the desired curve along which a heavy point falls from A to B in the shortest time, and let C and D be two points on it as close together as we like. Then the segment of arc CED is among all segments of arc with C and D as end points the segment that a heavy point falling from A traverses in the shortest time. Indeed, if another segment of arc CFD were traversed in a shorter time, then the point would move along ACFDB in a shorter time than along ACEDB, which is contrary to our supposition.
Jacob's lemma generalizes to any case where the curve that you are trying to find is characterized by satisfying a global extremum condition.
If the curve satisfies the extremum condition globally then every subsection of the trajectory satisfies the extremum condition too, down to infinitissimally short subsections.
It follows that if the curve satisfies an extremum condition then a differential equation must exist that has the sought after curve as its solution.
Back in the time of Jacob and Johann Bernoulli there was no systematic way of identifying the corresponding differential equation.
This problem was addressed by Euler and Lagrange, hence their systematic approach is called the 'Euler-Lagrange equation'.
I will come back to the Euler-Lagrange equation further down in this answer.
Comparison: numerical analysis
I recommend the following way to see the connection between variational calculus and differential calculus: implementation in numerical analysis.
There is a Java applet created by Todd Timberlake that offers a numerical implementation of Hamilton's stationary action
For differential calculus the simplest algorithm is Euler integration. Time is divided in small increments, the calculation proceeds increment by increment.
The numerical analysis implementation of variational calculus is as follows: time is divided in small increments. Then there is an iterative calculation: In each step the algorithm evaluates the middle point of a triplet of points, advancing one time increment each step. In each step the middle point is adjusted relative to its directly adjacent points, to make it satisfy the extremum condition locally:
$t_0$, $t_1$, $t_2$
$t_1$, $t_2$, $t_3$
$t_2$, $t_3$, $t_4$
And so on.
When the last point is reached the iteration start again with the first point
The simplest seed for the iterative calculation is a flat line. The iteration can start with any seed.
Over each iteration the trial trajectory shifts towards the true trajectory. Over a sufficiently large number of iterations the trial trajectory converges to the true trajectory to within any desired level of computational accuracy. So: the numerical analysis implementation arrives at satisfying the extremum condition globally by iteration of evaluating the extremum condition locally.
And of course: the smaller the time increments the higher the level of computational accuracy.
It is occasionally claimed: variational calculus is unique in that it is based on evaluating whether a global condition is satisfied. However, that claim is untenable; there is only one way to implement the numerical analysis: by division in small time increments, and the global evaluation is built up by way of iteration of local evaluation.
Is Hamilton's stationary action a form of knowing the future?
I surmise that the underlying thought of your question is: "Is a particle able to know the future in advance?"
Well, it's not the case that Hamilton's stationary action as a calculation strategy implies "knowing the future in advance".
Hamilton's stationary action is a form of extrapolation.
As we know: the simplest form of extrapolation is for motion along a straight line: The distance between point A and point B is X units of distance. How fast must you travel to make it to point B to arrive in Y units of time?
Next level up is to make the trajectory curvilinear: a projectile deflected by gravity. The distance between point A and point B is X units of distance. A projectile is fired at an angle to the horizontal of $\alpha$ degrees. At what velocity must the projectile be fired in order to make it to point B in Y units of time?
The trajectory involves deflection now, but we can still solve for the required velocity, and we do not find in in the least surprising that we can still still solve the problem.
(It so happens we know that with gravity the trajectory follows a parabola, but not knowing that in advance does not present a problem: solving for that parabola will be an organic part of the overal course of the solution.)
Either you get the velocity as initial information, and you use that to calculate the expected duration, or you get the expected duration as initial information, and you use that to calculate required velocity.
Hamilton's stationary action as a calculation strategy is no different from the above. You don't know the shape of the trajectory in advance, but you solve for that organically in the overall process of solving the problem.
the Euler-Lagrange equation
In this section I examine the strategy of the derivation of the Euler-Lagrange equation.
Mathematicians always push for more generalisation: the Euler-Lagrange equation is the general strategy for the general variational problem.
For transparency I will narrow down to application of the Euler-Lagrange equation in dynamics, more specifically classical dynamics.
For dynamics the Euler-Lagrange equation expresses the derivative with respect to position of the Action. In dynamics one might expect to evaluate derivative with respect to time, but the EL-equation expresses derivative with respect to position.
The potential energy is a function of the trajectory, and the kinetic energy is a function of the trajectory. So you have the freedom to take the derivative with respect to position instead of the derivative with respect to time.
The starting point is Hamilton's stationary action.
The first step is to express the derivative of Hamilton's action with respect to the variation.
In the course of the derivation of the EL-equation it is demonstrated that the derivative with respect to position and the derivative with respect to variation are the same, since the variation is variation of position.
So, while the starting point was derivative with respect to variation, the derivation then transforms that in to derivative with respect to position.
On how to derive Hamilton's stationary action
For any curve that satisfies a global extremum condition we have that a differential equation exists that has the sought after curve as its solution.
We can take that in reverse, and transform a problem in differential calculus to a problem in variational calculus.
We start with the Work-Energy theorem.
$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (1) $$
I noticed that this theorem is often referred to as Work-Energy principle. It is a theorem in the following sense: if we take the derivative with respect to position we recover $F=ma$ immediately.
Simplify to initial position coordinate zero and initial velocity zero
$$ \int F \ ds = \tfrac{1}{2}mv^2 \qquad (2) $$
Derivative with respect to position:
$$ \frac{\int F \ ds}{ds} = \frac{\tfrac{1}{2}mv^2}{ds} \qquad (2) $$
$$ F = ma \qquad (3) $$
From here on I will refer to the negative of the integral of force over distance as the 'potential energy'.
From the work-energy theorem it follows:
$$ \Delta(E_k) = \Delta(-E_p) \qquad (4) $$
Coming from the work-energy theorem (4) is fundamentally limited in scope as compared to the Principle of conservation of energy. The principle is unconditional, but the work-energy theorem is applicable only if there is a defined integral of the force over spatial coordinate.
(4) is valid down to infinitisimal increments.
$$ d(E_k) = d(-E_p) \qquad (5) $$
The derivative with respect to position:
$$ \frac{d(E_k)}{ds} = \frac{d(-E_p)}{ds} \quad (6) $$
About integration:
Integration is a linear operation. If you take a curve, and you multiply that curve with a factor $a$ then the integral of that curve will be larger by that factor $a$.
$$ \int ax \ dx = a \int x \ dx $$
We take the property that is expressed in (6) and state it as integrals of the respective energies:
$$ \frac{d(\int E_k \ dt)}{ds} = \frac{d(\int -E_p \ dt)}{ds} \quad (7) $$
The property expressed by (6) propagates to (7) because integration is a linear operation.
We know from the derivation of the Euler-Lagrange equation that the derivative with respect to position and the derivative with respect to variation are the same, hence (7) is mathematically equivalent to stating Hamilton's stationary action.
