Where does a U-turning car get its energy from? When you're flying a spaceship (away from large bodies with gravity), and suddenly realize you've left the oven on and need to turn around, you need to burn $2E_k$ worth of fuel to do it (where $E_k$ is your kinetic energy): firstly to break to a stop, and secondly to accelerate in the opposite direction back to your previous speed.
However, when instead driving a car, you can just turn around (on a large enough radius so that you don't skid) when not in gear, without hitting the acceleration pedal at all (or even with your engine off). Sure, you'll lose some speed, but you can manage to somehow bypass this large energy requirement while changing your momentum to the entirely opposite direction.
Where does this extra energy come from when you turn a car around? What happens to your original energy? Mind you, it appears you only have $E_k = \frac{mv^2}{2}$ energy to begin with, but need twice that much to turn around in a vacuum.
I'm guessing tires and asphalt play a huge role here, but it seems quite counter-intuitive that they can somehow produce twice the energy you originally had. What provides that energy?
 A: You can turn your spaceship around without spending $2E_k$. Just go half-way round the Moon or some other planet. The point is that as long as the force changing your direction is always at  right-angles to your velocity vector, it does not  change your KE.  For  the cornering car, the static friction force on the tyres is at right angles to the car's velocity.
A: 
When you're flying a spaceship (away from large bodies with gravity), and suddenly realize you've left the oven on and need to turn around, you need to burn 2Ek worth of fuel to do it (where Ek is your kinetic energy): firstly to break to a stop, and secondly to accelerate in the opposite direction back to your previous speed.

This is not true for several reasons.
First, rockets work by conservation of momentum between the rocket and the exhaust, so the energy that you must use if you are using your engines to execute this maneuver will also include all of the energy that you put into the rocket exhaust. If you track it carefully, you will find in fact that all of the KE goes into the exhaust and none into the rocket on average.
To put some specific math behind this statement suppose that the rocket works by firing one instantaneous burst of exhaust, all at the same velocity (it is possible to do a similar analysis using the rocket equation, but that adds some complexity that obscures the point). The rocket has mass $M$ and the exhaust has mass $m$ and leaves the rocket at a relative velocity of $v_e$. The initial KE of the rocket is $E_k=\frac{1}{2}M v^2$, and the energy from the fuel burned is $E_f$, so we can write conservation of energy and momentum as $$\frac{1}{2}M v^2 + \frac{1}{2}m v^2 + E_f = \frac{1}{2}M (-v)^2 + \frac{1}{2}m (v+v_e)^2$$$$M v + m v = M (-v) + m (v+v_e)$$ From this it is straightforward to show that $$E_f=\frac{1}{2}m (v+v_e)^2 - \frac{1}{2} m v^2$$ meaning that all of the energy from the fuel goes into increasing the KE of $m$ with none left for changing the KE of $M$. It is also straightforward to show that $$E_f=2 M v^2 + M v v_e = \left(4 + 2\frac{v_e}{v} \right) E_k \ne 2E_k$$ so, not only is the energy burned in the fuel never equal to $2E_k$, but also for large $v_e$ it can be arbitrarily large compared to $2E_k$
Second, you can do it for "free" if you happen to have a planet or star you can use. Although you excluded that in your question, the fact that it is possible shows that the energy "requirement" is not actually a physical requirement but is simply due to using an inefficient method.

Where does this extra energy come from when you turn a car around?

There is no extra energy. You are just using a more efficient method. Similarly, if you hang a helicopter from a rope it hovers without expending any energy, but if you have it hover with its rotors it expends a lot of energy.
