Why does something on an inclined plane move forward at all? We just started studying about the inclined plane and vectors in motion, but i don't understand one thing: Why on earth does the object on the inclined plane move forward (I.e in the direction where the inclined plane is 'falling') at all?
I understand that the force due to gravity is (almost) perfectly downward. I can also understand that since the object sits on the inclined plane, there should be an equal and opposite force on the object (Owing to the fact that when the object pushes down on the inclined plane, it gets slightly closer to the inclined plane, which increases electrostatic repulsion and makes the atoms that are part of the upper layer of the inclined plane in turn push the object 'up'), but that should be (almost) perfectly upwards [The almost part owing to the curvature of the earth would be too small to make a difference].
So in this case, how on earth does the object on the inclined plane gain any 'forward' velocity at all?
My question is, when you put an object on an inclined plane, how the heck does it get horizontal velocity?
 A: Have you ever walked down a mountain slope? Did you ever feel that you climb down faster than you normally walk? 
How did you get any horizontal velocity at all?

This image is a classic textbook illustration of an object on an inclined plane. 
mg is the weight of the object. It is acting vertically downwards. 
But you can imagine that this vertical force is made up of two forces - One acting perpendicular to the wedge and other parallel to the slant of the inclined plane. 
A little trigonometry reveals that these forces are mg cos$\theta$ and mg sin$\theta$ respectively. 
Now you need to forget about the original vertical force mg. Consider you have these two mentioned forces only.
mgcos$\theta$ is acting perpendicular to the incline of the wedge. This will not provide the object with any horizontal velocity. It is responsible for the reaction force N by the wedge. The frictional force f is proportional to this component.
mgsin$\theta$ is parallel to the incline. This force will try push the object down the incline. But behold! f is opposing this force!
If mgsin$\theta$ is greater than maximum possible value of friction, the object will experience a force mgsin$\theta$ - f, and this force will be responsible for making the object move down the slope.
That's how you get a horizontal force.
A: 
I understand that the force due to gravity is (almost) perfectly
  downward. I can also understand that since the object sits on the
  inclined plane, there should be an equal and opposite force on the
  object

No.  If we are considering the pull of the earth on the object as the force in question, then the "equal and opposite" force is the one that the object exerts on the earth, pulling it upward.  

(Owing to the fact that when the object pushes down on the inclined
  plane, it gets slightly closer to the inclined plane, which increases
  electrostatic repulsion and makes the atoms that are part of the upper
  layer of the inclined plane in turn push the object 'up'), but that
  should be (almost) perfectly upwards

The electrostatic repulsion does push the object, but the net direction is perpendicular to the plane, not upward.  This is the normal force.  
When this force is not vertical (the plane is not horizontal), and there is no friction, then the forces are unbalanced and the object accelerates with some horizontal component.
