You have a fundamental misunderstanding about the relationship between velocity and acceleration.
- Velocity is the change in distance over time ($m/s$)
- Acceleration is the change in velocity over time ($m/s^2$)
Perhaps answering the questions you posed will help you understand this relationship better:
1) Would it mean the object is not accelerating but still in motion from the left over force? (example; after car stops accelerating it is just rolling)
In this case there seems like there would be velocity just slowing down.
Yes, if acceleration dropped to $0$ the object would still be in motion, but not because of some 'left over force', because the velocity of the object would not longer be changing. Look back to the relationship I described above, acceleration is the change in velocity over time, not just the velocity over time. If acceleration is $0$, the velocity is not changing. If the velocity is constant ($0$ acceleration) then the object will continue without slowing down or speeding up.
2) Or does this graph describe an object who is accelerating @ 0s-10s and stopping all motion @ 10s-15s? In this case mathematically makes sense because if you solve for velocity V = a*t you get 0 m/s
The graph describes an object which is accelerating from $0s$ to $10s$, and not accelerating from $10s$ to $15s$. As in the last question, the fact the the object is not accelerating does not mean that the velocity is $0$, it means that the velocity remains constant at whatever value it was just before acceleration became $0$.
Additionally, $V \ne a*t$, because the relationship is not between velocity and acceleration, but change in velocity and acceleration. The proper formula would be $\Delta V = a*\Delta t$, where $\Delta t$ would in this case be $15s - 10s = 5s$ This would mean that $\Delta V = (0m/s^2)*(5s) = 0$, so velocity would not change, but would remain constant.