What relative motion tells me about a motorbike shooting out from a bus? I imagined this situation:
Suppose I'm on a motorbike inside a long bus that has a constant velocity v on a street.
I start accelerating (in the direction opposite to the motion of the bus) till reaching a velocity u  that is less than v (inside this long bus).
The rear of the bus (not a very safe one) is open, and I bravely go past it at my velocity u, yelling "Mom, Watch! Watch!".
Now, since I really don't want to make it too complicated beacause I can't handle it,
let's suppose the bus and the street are on the same level, and we are in perfect vacuum,there's no friction and  I never accelerate beyond my u speed!
A friend of mine is standing still on the ground:
What will he see ? Will I end up moving in the direction of the bus with a speed (v-u),according to him? Will I keep moving forever?
If there's any problem with the question please address it in the comments and I'll edit it!
 A: Answer for $u$ is relative to the ground
At your level of simplification (and assuming reasonable, not relativistic velocities), it really makes no difference whether you are inside the bus or just driving next to it (calling the time that you are on the side of the bus "inside the bus"). So relatively to the ground you are always moving with velocity $\mathbf{u}$ and the bus always with velocity $\mathbf{v}$.
Depending on the size and direction of these velocities you can end up with different situations:


*

*If $\mathbf{u}$ and $\mathbf{v}$ have opposite signs, you are moving in opposite direction to the bus and your relative velocity is $\mathbf{v}-\mathbf{u}$, its absolute value is $|\mathbf{v}|+|\mathbf{u}|$. Here and in the following I make use of the fact that the motion is one-dimensional.

*If $\mathbf{u}$ and $\mathbf{v}$ have the same sign you are moving in the same direction. $|\mathbf{u}|<|\mathbf{v}|$ has to hold as otherwise you would never be able to exit the bus at its end. The relative velocity is $\mathbf{v}-\mathbf{u}$, its absolute value is $|\mathbf{u}-\mathbf{v}|$.



Will it last forever?

Newton's law is telling us that bodies move at constant velocity until a force is acting on them. I don't see any forces specified in your question, so yes, this will last forever.
Answer for $u$ is the velocity relative to the bus
If $u$ is the velocity relative to the bus the velocity of the motorbike relative to the ground is: $\mathbf{v}+\mathbf{u}$ in vector notation or $||\mathbf{v}|-|\mathbf{u}||$ in absolute value because $\mathbf{u}$ and $\mathbf{v}$ have opposite sign/point in opposite directions. With your assumption, $|\mathbf{u}|<|\mathbf{v}|$, the bike ends up moving in the same direction as the bus. I realize that this is not realistic with the wheels of the bike spinning in the opposite direction. However this is just an artifact of the simplified model. If you wanted to make it more realistic you would have to consider friction which would make this problem a lot more complicated.
In the opposite case, $|\mathbf{u}|>|\mathbf{v}|$ the bike would end up going in the opposite direction to the bus. In either case the speed of the bike with respect to the ground is given by the difference of speeds of the bus and the bike with respect to the bus.
As above, in this simple model, there is no difference whether the bike is actually inside the bus or not. And as above, this will last forever (within this simple model).
You can transform between reference frames like these using Galilean transformations
