Friction and Work in Inertial Reference Frames 
Say a train is moving in the positive $x$-direction at 100 meters per second with respect to the ground frame. Now let's say someone is pushing a large box in one of the cars of that train in the negative $x$-direction and is applying sufficient force to keep the box moving at a constant 5 meters per second in reference frame of the train car. 

My intuition keeps telling me that this is like any other problem where the ground frame observer would witness the force being applied over a longer distance compared to the pusher/observer in the train frame, so work done is greater in ground frame,  but isn't the energy dissipated as heat independent of reference frame? 
Furthermore, the ground observer would see the box moving forward at 95 meters per second as the train moves forward at 100 meters per second. Correct? The person pushing is doing work and the friction is doing equal work in the opposite direction. From the ground frame we'd see much more work done but how do we account for the dissipated energy such as heat and vibration? I feel like from the ground frame you'd see enough heat to start a fire.  I hope I have articulated my confusion clearly enough.  
 A: Sliding friction is a bit odd. With most interactions you can use the power equation quite easily: $P=F\cdot v$. With sliding friction it is a little confusing because of the issue about which $v$ should be used since each surface has a different $v$. It turns out that you use both and the difference is the power that goes into thermal energy. 
So assuming $F=10\:N$ then in the train’s frame the power on the box’s side of the interface is $P_{box}=F\cdot-v= -50 \: W$ and the power on the train’s side of the interface is $P_{train}=-F\cdot 0=0 \:W$. So 50 W is going into the interface and not coming out, this is the energy that increases the temperature. 
In the ground frame the train’s velocity is $u=100 \: m/s$ so $P_{box}=F\cdot (u-v)=950 \: W$. Note, the sign of this power has changed, so this is energy that is going out of the interface and into the box, which is the opposite direction as previously. However, now $P_{train}=-F\cdot u=-1000 \: W$ meaning that 1000 W goes into the interface from the train side. Note again that 50 W goes into the interface and does not come out. 
So the same 50 W of “heating” occurs in both frames. The explanation of where that power comes from is different, but all of the observable measurements are the same. 
The nice thing about this approach is that there is no special reference frame. Any reference frame gives the same result with a simple systematic use of the standard equations. 
A: The heat produced by kinetic friction corresponds to the distance an object moves along a frictional surface, not the distance it moves in some arbitrary reference frame. In other words, the velocity used in heat from friction calculations is the velocity of the object with respect to the surface. In this example for the ground reference frame, the velocity of the box is 95 m/s and the velocity of the floor of the train is 100 m/s. Therefore, one would still find the sliding velocity between the box and the floor to be 95 - 100 = 5 m/s, so the heat produced by friction would be the same. Note that this is not the same as work done by friction.
A: 
My intuition keeps telling me that this is like any other problem where the ground frame observer would witness the force being applied over a longer distance compared to the pusher/observer in the train frame>

This is true. In each frame the net work is $0$, but the work done by each force depends on the frame. In fact, the work done by each force has different signs compared between your frames.
However, when it comes to how much energy is dissipated and lost due to heat, then only the relative motion between the box and the train matters (Dale's answer gives a proof of this, where it is shown that the power dissipated depends on the relative motion between the surfaces), and so it would be best to work in the frame of the train.
This is a good question, and causes us to really think about work and friction. We usually equate the work done by friction to the energy lost due to heat simply because most friction problems are done assuming we are at rest relative to the surface the object is sliding on. However, as your example has shown, this cannot always be the case.

A simple counter-example showing that the work done by friction does not always give the energy "lost to heat". Let's say I hold a box stationary while a conveyor belt moves underneath it. According to me, the box is not moving, so the work done by friction (and myself) must be $0$, yet I'm sure we would all agree that the box would start to heat up due to friction. To find how much energy is being lost to heat, we could move to a reference frame moving along with the belt so that the belt is at rest. Then the box would be moving relative to us, and any motion we see of the box is motion relative to the belt. Then the work done by friction would be the energy lost as heat.
