Why does adding a term $5f'(t)$ to $5f''(t)+10f(t)=0$ cause damping?

If $f(t)$ stands for the displacement of the oscillator from its equilibrium, then the first derivative $f'(t)$ will be the velocity and the second derivative $f''(t)$ the acceleration (which, according to Newton's $2^{nd}$ law of motion, is always proportional to the resultant force exerted on the oscillator).

So, your first equation $5f''(t)+10f(t)=0$ says that: $\sum F(t)=-kx(t)$, ($k$:constant) which is Hooke's law i.e. the necessary and sufficient condition for a free (undamped) oscillation.

On the other hand, your second equation $5f''(t)+5f'(t)+10f(t)=0$ says that: $\sum F=-kx(t)-bu(t)$, ($k,b$:constant) i.e. there is a component of the net force proportional to the velocity: $-bu(t)$, which is exactly the damping force. (Usually frictional forces, air resistance and more generally various resistances to motion are modelled by summands proportional to the velocity or to some power of it).

Let's write the differential equation like this: $$f''(x)=-2f(x)-f'(x).$$ The acceleration of the oscillator is equal to $-2$ times its position minus its velocity. According to the initial conditions, this means that the initial velocity is positive, position is zero, so the initial acceleration is negative. Since velocity and acceleration have opposite sign, the oscillator is slowing down. In the original system, the acceleration at the beginning was zero, so when we add the damping term, the oscillator slows down more abruptly than before. This means that the first maximum of position is a little less with damping than otherwise. Now the oscillator enters the speed up phase. But since the Oscillator's maximum is not as extreme as before, its acceleration is smaller in this phase, so when it reaches its maximum velocity, its speed is also smaller. The oscillator is losing energy. The pattern continues forever. The oscillator loses energy and decreases in amplitude exponentially.