If we ignore energy and talk about force, Why do objects bounce lower when they hit a softer surface? If an object hits a soft surface it will bounce lower compared to the object hitting a hard surface, isn't the impulse in the first case equal to the impulse in the second case, so why does the object bounce lower when it hits a softer surface? 
 A: When an object bounces off of something, the bounce is not instantaneous, and in fact goes through a sequence of changes.
First, the object slows down to a stop, and both the object and the surface are deformed. If they are both hard, then the deformation will be very small, but it will still exist.
Second, both the object and the surface reverse their deformations, speeding the object back up in the reverse direction.
The difference that makes a softer surface produce lower bounces is that soft surfaces stop that second stage early, leaving the surface still partially deformed. This reduces the time of contact, and thus the impulse of the bounce. The energy lost to this is converted primarily to heat.
More spring-like surfaces also do it more gradually, leaving some of the un-deformation still in progress when the object loses contact with the surface. This reduces the force during the time of contact, and thus the impulse of the bounce. The energy lost to this is converted to more macro-scale vibration, which may even be readily visible.
Illustrating with an example:
A ball has 1000 kg * m/s momentum when it first touches a soft surface. The ball and surface each exert 100 N force on each other as long as they are in contact.
The first stage, the "impact", lasts 10 seconds. During the impact, the ball slows to a stop and deforms slightly, and the surface deforms a lot. Both the ball and surface cause an impulse of 1000 N * s on each other, during this stage by itself.
The second stage, the "rebound", lasts only 5 seconds. During the rebound, the ball accelerates away and un-deforms, and the surface un-deforms only partially. Both the ball and surface cause an impulse of 500 N * s on each other, during this stage by itself.
The end result is that the ball bounces off with only half the speed it started with. For a hard surface in the same scenario, the rebound would instead last almost as long as the impact, about 10 seconds, the surface would un-deform all the way, and the ball would bounce with almost as much speed as it started with.
P.S. Notes on concepts and units:
Momentum: The product of mass and velocity. It has unit kilograms * meters / second, or kg * m/s. This unit is equivalent to N * s.
Force: How hard you're pushing on something. Its unit is Newtons, or N.
Impulse: The product of force and time, and also the change in momentum. It has unit Newtons * seconds, or N * s. This unit is equivalent to kg * m/s.
Newton's third law applies to force, and indirectly to impulse. It does not apply to momentum.
In typical everyday collisions, gravity applied over some previous time span may be the source of some of the momentum involved, but is inconsequential for the collision itself because it is utterly dwarfed by the forces exerted by the objects on each other. The collision between a ball and a hard floor takes something like a millisecond to finish, but produces an impulse large enough to reverse - not just stop - momentum that took around 500 milliseconds for gravity to build up.
A: Let's break down all the forces involved. Note: I've surely missed a bunch, and this got even more complicated than I was expecting when I started writing it. 
Stage 1: Ball is in your hand before you drop it. 


*

*Force 1: Gravity on the ball (Opposite force is the ball pulling on the earth, but this is inconsequential to this problem)

*Force 2: Depending on how you're holding the ball, either normal force or gravity. The sum of both which will equal the force of gravity on the ball (if they didn't the ball would move) 


Stage 2: Ball is let go, and begins falling. 


*

*Force 1: Gravity on the ball 

*Force 2: Air resistance on the ball

*Force 3: Force applied by the ball on the surrounding air to move it out of the way so that it can take that new place. Force 3 == Force 2 

*Net force down as acceleration is observed 


Stage 3: ball begins contact with ground. 


*

*Force 1: Gravity on ball.

*Force 2: Normal force applied by ground to ball.

*Force 3: Force applied by ball to ground. Force 3 == Force 2 

*Force 4: Anti-compression force. Couldn't think of a better name here, but as you compress a ball, it has a tendency to return to its ball shape. This force gets higher and higher as you compress the ball. This happens with the ground as well, but at a different "rate". I'll get into what I mean by rate in a bit 

*Force 5: Going to group a lot of things here, but the ball is pushing against the ground, which is rippling vibrations through the ground and air. 

*Net force: Well, very very very small amount of net force down at part of the ball that has impacted the ground. The top of the ball has a much higher net force down at this point. 


Stage 4: Compression Equilibrium. This is the point at which the ball stops accelerating downward. This is a very complicated system to pin down, but this is the point where the anti-compression force is equal to the normal force applied to the ground. I.E the moment before the ball starts bouncing back 


*

*Force 1: Gravity 

*Force 2: Anti-compression force. The ball is now returning itself to a ball. As a consequence of this the ball will apply a force to the ground. 

*Force 3: The ground also pushes back against this anti-compression force to apply an upward force to the ball

*Force 4: Air resistance above the ball fighting against the anti-compression. 

*Force 5: The ground may or may not be bouncing back at this point. It depends on the hardness/softness/elasticity of the ground relative to the ball. This will greatly change how much of the normal force is applied to the ball at this step. This is what I was referring to by a different rate. There's no guarantee that the bounce back from the ground and the anti-compression of the ball will be in sync. 

*Force 6: More vibrations through the ground 

*Net force: Up (or at least up the moment after equilibrium. There is a moment of time where the net force on the ball is actually 0!)


Stage 6: Ball leaving the ground 


*

*Force 1: Gravity 

*Force 2: Air resistance, the ball is no longer impacted by the ground anymore now 


As you can see from these forces, a lot of force has gone to places that isn't propelling the ball back to its original place. 
But regardless of that, look how complicated this analysis is. There's too many little interactions between forces to make good sense of what's happening. This is precisely why we use energy to reason about these things. Because we have a conservation of energy law that greatly simplifies all calculation, and we have a solid understanding of kinetic energy, and heat. Our understanding of elastic collisions is basically just experimentation, then building theories around the system as a whole, rather than trying to derive it from more primitive/fundamental physics. 
The most important thing here is that there is no such equivalent with forces. The closest thing to a conservation of force is that forces must have an equal an opposite force at all times. But do not misunderstand that, that tells you nothing in this example, as you can see from all the forces listed, they all have opposite forces, but the net effect is still lost energy. 
A: You need to be careful as to exactly what you mean by the terms "softer" and "harder". It is rather the degree of inelasticity of the collision that determines how high the object bounces. 
For example, you can have a spring with a low spring constant that you might consider "softer" than a spring with a higher spring constant which is "stiffer". If the two springs are ideal (no friction losses) an object bouncing off each of them will reach the same height because the collisions would be considered perfectly elastic. (Assumes the object itself is perfectly rigid).
However, at the macroscopic level all real collisions are inelastic. Kinetic energy will be lost due to friction associated with the inelastic deformation of the colliding objects. Since most softer surfaces will undergo more deformation, it stands to reason that the collision with such surfaces will be more inelastic than harder surfaces as they potentially undergo more deformation. 
The degree of inelasticity of a collision is reflected in the coefficient of restitution (COR) of the colliding objects. That is the ratio of the final to initial relative velocity between two two objects after they collide, and is a number less than 1.  All other things being equal the COR will likely be lower in a collision with a softer surface depending on how much it deforms and the degree to which the deformation is permanent.
Shouldn't there be an equal force in magnitude and opposite in direction when the object hits the surface in both cases? therefore in both case shouldn't the object bounce back to the same height?
Per Newton's third law the force the surface exerts on the object is equal and opposite to the force the object exerts on the surface. But the magnitude of the equal and opposite forces will not be the same when the object initially contacts the surface as when it comes off the surface after penetrating the surface. That's because energy is lost as heat when the object penetrates the surface. That means there is less kinetic energy of the object when it comes off the surface then when it initially hit the surface. Less kinetic energy means less velocity and a lower rebound height.  
Hope this helps.
A: An inelastic collision is an example of friction: One of the objects, in your example the surface, will plastically deform while its innards are rubbing on each other, producing heat. If there is no elasticity at all, say with a heap of soft sand, no energy other than heat is stored in the ground during the impact. If the ball hitting it is sufficiently hard, e.g. made of steel, no significant amount of energy is stored in it on impact, either. Because friction only exerts a force on the moving objects as long as they move, the steel ball will simply form a dent in the sand (or lump of clay, or whatever) until its momentum and energy is used up resp. transferred, and then sit there, with no repellent force acting upon it. That's why it does not jump back up. Depending on the mass of the second object some or most of the original kinetic energy has been converted into heat, while the overall momentum of the ball-surface system is of course unchanged: The collision consisted of only inner forces which do not change the momentum of the overall system.
Note that when the collision partner is the Earth, like when we bounce a Super Ball off the tarmac or throw it into a heap of sand, we must be careful with momentum considerations: Because Earth's mass is so huge. It is not obvious that momentum is conserved during the fall or during the bounce because the ball accelerates and even completely inverts direction seemingly violating the law that momentum be conserved.
But it is important to understand that the attraction between Super Ball and Earth, as well as all forces during a collision, are entirely mutual. The Earth starts moving towards the ball when the ball falls; both experience exactly the same change of momentum in exactly the opposite directions (which must be so because the attraction is an inner force and cannot change the overall momentum...). It's just that the Earth weighs 6 million billion billion kilograms as opposed the the Super Ball's 0.02, so that when the ball moves with a speed of 5 m/s towards the Earth before impact, the Earth moves with only 1/60 million trillion trillionth m/s towards the ball which is only noticeable if you look really close. Joke aside, this effect is measurable or even pronounced with moons orbiting planets, or planets orbiting the sun: Being equally attracted by the orbiting bodies, the central body "wobbles" sometimes so much that it's entirely outside the system's center of gravity.
A: The collision with the surface lasts much longer than it would if the surface was hard. The force applied on the object by the surface is given by the formula: $\sum F=\frac{\Delta p}{\Delta t}$ where $\Delta p$ is the change of momentum of the object ($p_{final}-p_{initial}$ where $p_{initial}$ is $mu$ and $p_{final}$ is $0$). From this formula it is obvious that when $\Delta t$ is getting smaller the total force is increasing. 
A: Why do you assume that the impulse is the same in both cases? Impulse is equal to the change of momentum, so clearly it is not the same. There is an equal and opposite force in both cases, but these forces are not the same in one case as in the other (indeed the meaning of "soft" is a smaller equal and opposite force during collision as compared to "hard"). Also the time during which the force acts is different, so there really is no reason to think the impulse would be the same. 
You really can only think of the difference in terms of the work done by the equal and opposite forces leading to energy being transformed into heat.
A: No, the final speed is lower with a softer surface, so the impulse is lower.
If you throw a ball at a hard surface, the ball will not stop instantly. It will compress as it slows down. The energy of that compression will then be released, pushing the ball back in the direction opposite of the original motion. In the theoretical case of perfect elasticity, all of the energy will be retained by the ball, and the final speed will be the same as the original.
When a ball hits a soft surface, some of the energy is absorbed by the surface, rather than the ball, deforming. If the surface is elastic, then that energy will be restored to the ball. But if the surface is inelastic, that energy will be dissipated into heat. The ball will leave with less energy, and thus less speed. 
