Why do we weigh less when falling? 
I don't want to go to science world to find out because it would be a long round-trip.
I understand that acceleration/deceleration would effect the weight and I can also imagine that someone at terminal velocity would weigh nothing but I can't get my understanding in terms of forces and how that would effect the weight. For example, what would a formula relating mass, speed (in direction of gravity), gravitational force and weight look like?
 A: As an alternative example, think about how it feels to drive a car.


*

*When the car is standing still, you will sink into the seat and feel the seat pressing against you (= normal weight).

*When you press the throttle and the car accelerates, you will be pressed further back into the seat and will in turn feel the seat pressing harder against you (= increased weight).

*When the car has reached cruising speed and is not accelerating any longer, you will not get further pressed into the seat any longer (= normal weight again). There are still forces acting on the car (engine is driving it, wind resistance and friction are slowing it), but they are balanced and results in no acceleration.

*When you press the brake and the car decelerates, you will be lifted out of the seat, which will in turn lessen it's pressure on you (= decreased weight).


In other words, your weight will only change when the car is accelerating/decelerating.
A: 
"I understand that acceleration/deceleration would effect the weight and I can also imagine that someone at terminal velocity would weigh nothing"

It looks like you think that only in terminal velocity we do not weight, and that's wrong. We measure weight by the normal force applied by the ground on us. That means that there must be a normal force. In the elevator, it first accelerates down with an acceleration $a<g$. 
In terms of forces, you must accelerate at the same rate as the elevator so $ma_T=ma-mg$ the normal force must be smaller than just $mg$, and so you weigh less. In the terminal velocity of the elevator the total forces must be 0 as your speed is constant, so the normal must be again $mg$.
In free fall, the only force that acts on you if the gravity, and so you weigh nothing, in space for example, in the ISS you see them floating no because there's no gravity (gravity at the altitude of the ISS is almost the same as in the surface), but because gravity is the only force acting on the whole system, so they weigh 0.
A: You could think of the weight recorded by the weighing machine as the force exerted by our mass on the weighing machine. I'm using mass to mean inertial mass and not weight. 
You're probably aware that the force $F$ exerted on a mass $m$ is related to the acceleration $a$ as $$F = ma$$
In context of gravity, the acceleration is provided by the gravity itself, and the 'force exerted' is called the weight.
In free fall (and to a lesser extent in a downward moving elevator), both you and the weighing scale are falling toward earth with the same acceleration. Therefore there is no "extra" acceleration that you posses by which you can exert a force on the weighing scale and hence register a non-zero weight.
A: Your weight is the force you apply to the floor that supports you. If you stand on the ground and your mass is $m = 80$ kg, then the ground is feeling a force of $ m \times g = 80$ kg $\times$ $9.8$ m/s$^2= 784$ N.
If you're in an elevator that's accelerating downwards with $a = 1.0$ m/s$^2$, then the floor of the elevator feels a force of $m \times (g - a) = 80$ kg $\times$ $(9.8$ m/s$^2 - 1.0$ m/s$^2) = 704$ N. In other words, your weight on the elevator is as if your mass were $704$ N$ / 9.8$ m/s$^2 = 71.8$ kg instead of $80$ kg.
If the elevator accelerates downwards with $a = 9.8$ m/s$^2$ (that is $a = g$, it's in free fall), then the floor feels a force of $m \times (a - g) = 80$ kg $\times$ $0$ m/s$^2 = 0$ N, so you're weightless.
A: Assume you are in a lift, and you weigh yourself with a balance. Your weight will be the normal force between balance and your body.
Let the normal force be $N$, the gravitational force be $m g$ (where $g$ is a constant) and the lift moves downward as positive,
Net force: $m g - N = m a \implies N = m \left(g - a \right)$
when it accelerates downward, $a$ (acceleration) will increase, which means $g - a$ will decrease where $m$ (mass) is constant, and $N$ will decrease, hence your weight.
