How do reaction engines create a force against the rocket? I'm fairly new to physics, so please forgive me if I lack basic understanding. 
I'm wondering how rocket engines actually work. I know the answer of "actio est reactio", but I feel like this is a very abstract and unclear explanation because it doesnt touch whats actually going on. Here is my issue:
As I understand, the engine generates an explosion which throws out gas particles into one direction (say down) and thus creates an inverted force that pushes the rocket into the other direction (say up). What I dont get is: How do the leaving particles interact with the rocket? They dont necessarily touch the rocket, at least not in a way that would make it go up, so how can they possibly have an effect? As far as I know, a force can either act through


*

*contact

*a field


Now I guess the gas particles dont create some sort of force field, so at some point there has to be a 'meaningful' contact.
I get that mathematically the force of gas particles that are coming out equals the force with which the rocket is accelerated, but I have a hard time understanding how particles that are already on their way out of the rocket are the cause of the acceleration.
My explanation would be that the rocket is flying for the same reason anything is 'flying' near an explosion: The explosives hit other objects and drag them with them (conservation of momentum). The same force applies in a rocket on all walls of the engine, and since the particles can escape "down", the only momentum the rocket gets is "up" (left and right cancel each other out).
But I run into issues with that explanation as well, because lets say we make the exit of the engine reaaally long, say a few hundred meters (thought experiment), and then close it up. now that means that at first the effect would be the same, rocket goes up, but once the explosion hits the other end of the engine, the rocket will be 'dragged' back again. seems strange to me.
please help me solve this issue. thanks!
 A: "The same force applies in a rocket on all walls of the engine, and since the particles can escape "down", the only momentum the rocket gets is "up""
That's it: the gas molecules collide with the walls of the combustion chamber, but there is a hole in one wall, so the force on that wall is less!
The net force on the rocket is equal and opposite to the momentum per unit time of the escaping molecules, as these don't hit the wall with the hole in it! 
"lets say we make the exit of the engine reaaally long, say a few hundred meters (thought experiment), and then close it up. now that means that at first the effect would be the same, rocket goes up,"
Yes.
"but once the explosion hits the other end of the engine, the rocket will be 'dragged' back again." 
No, but it will stop accelerating.
A: Here is another way to think of this problem which might help:
Imagine you are sitting on a skateboard with a cannon in your lap. This is a special cannon, consisting only of a short tube closed at one end. Inside the tube is a bowling ball that is held tightly in position against a squeezed spring sitting between the ball and the closed end. a trigger mechanism holds the system in its "cocked" state. 
you point the cannon away from you horizontally, put your feet up on the skateboard, and pull the trigger. 
The spring begins to expand violently, pushing the bowling ball backwards out the tube at great speed. But it is also pushing equally hard on the closed end of the tube, which recoils and kicks you right in the chest and thereby pushes you off in the opposite direction taken by the departing bowling ball. 
A: 
How do the leaving particles interact with the rocket? They dont
  necessarily touch the rocket, at least not in a way that would make it
  go up, so how can they possibly have an effect?

Consider the following simple situation.  

Initially the mass $m$ is travelling at the same speed as the hollow cylinder which is sealed on the left hand side.
The mass $m$ explodes into two equal mass fragments one moving to the left and the other to the right with speed $v$.
The mass moving to the right does not interact with the tube and carries with it momentum $\frac m 2\,v\, \hat i$ and it escapes out of the tube.  
The mass moving to the left carries with it momentum $\frac m 2\,v\, (-\hat i)=-\frac m 2\,v\, \hat i$, hits the end of the tube and rebounds with speed $v$ now having momentum $\frac m 2\,v\, \hat i$.
So the change in momentum of this mass is $\frac m 2\,v\, \hat i- (-\frac m 2\,v\, \hat i)= m\,v\, \hat i$.  
The change in momentum of the mass which was moving left is equal to minus the change in momentum of the cylinder, $-m\,v\, \hat i$, ie the cylinder is given an impulse to the left thus increasing the cylinders speed to the left.  
What happens in a rocket engine is much more complex than this and here is a diagram from the Wikipedia page Rocket engine showing the forces acting on the rocket combustion chamber on the left and a (de Laval) nozzle on the right both of which contribute to the forward propulsion of a rocket.


But I run into issues with that explanation as well, because lets say we make the exit of the engine reaaally long, say a few hundred meters (thought experiment), and then close it up. now that means that at first the effect would be the same, rocket goes up, but once the explosion hits the other end of the engine, the rocket will be 'dragged' back again. seems strange to me.

Consider the cylinder closed at both ends and the "explosion" occurring close to the left hand wall.
The mass moving left transfers momentum to the cylinder which now moves faster to the left.
Then the mass which was initially moving right rebounds from the right hand end of the cylinder and reduces its speed to what it was originally.  
Then the mass which originally was moving left and is now moving right hits the right hand end of the cylinder and the cylinder now moves right.  
etc
