Is kinetic energy a relative quantity? Will it make inconsistent equations when applying it to the conservation of energy equations? If the velocity is a relative quantity, will it make inconsistent equations when applying it to the conservation of energy equations?
For example:
In the train moving at $V$ relative to ground, there is an object moving at $v$ relative to the frame in the same direction the frame moves. Observer on the ground calculates the object kinetic energy as $\frac{1}{2}m(v+V)^2$. However, another observer on the frame calculates the energy as $\frac{1}{2}mv^2$. When each of these equations is plugged into the conservation of energy, they will result in 2 distinct results (I think).
 A: You should remain in one frame of reference when applying the law of conservation of energy. Then you should be fine.
A: Yes, kinetic energy is a relative quantity. As you might guess, this means that when you're using energy conservation, you have to stay within a single frame of reference; all that energy conservation tells you is that the amount of energy as measured in any one frame stays the same over time. You can't meaningfully compare the amount of energy measured in frame A (e.g. the ground) to the amount of energy measured in frame B (e.g. the train).
However, you can convert an amount of kinetic energy measured in one frame to another frame, if you know their relative velocity. If you're working at low speeds, the easy (approximate) way to do this is to just calculate the relative velocity, as you did. So if the train observer measures a kinetic energy $K = \frac{1}{2}mv^2$, the ground observer will measure a kinetic energy of $\frac{1}{2}m(v + V)^2$, or
$$K + \sqrt{2Km}V + \frac{1}{2}mV^2$$
(in one dimension).
If you get up to higher speeds, or you want an exact expression, you'll have to use the relativistic definition of energy. In special relativity, the kinetic energy is given by the difference between the total energy and the "rest energy,"
$$K = E - mc^2$$
One way to figure out the transformation rule is to use the fact that the total energy is part of a four-vector, along with the relativistic momentum,
$$\begin{pmatrix}E/c \\ p\end{pmatrix} = \begin{pmatrix}\gamma_v mc \\ \gamma_v mv\end{pmatrix}$$
where $\gamma_v = 1/\sqrt{1 - v^2/c^2}$. This four-vector transforms under the Lorentz transformation as you shift from one reference frame to another,
$$\begin{pmatrix}E/c \\ p\end{pmatrix}_\text{ground} = \begin{pmatrix}\gamma & \gamma\beta \\ \gamma\beta & \gamma\end{pmatrix}\begin{pmatrix}E/c \\ p\end{pmatrix}_\text{train}$$
(where $\beta = V/c$ and $\gamma = 1/\sqrt{1 - \beta^2}$), so the energy as observed from the ground would be given by
$$E_\text{ground} = \gamma(E_\text{train} + \beta c p_\text{train})$$
The kinetic energy is obtained by subtracting $mc^2$ from the total energy, so you'd get
$$K_\text{ground} = \gamma(E_\text{train} + \beta c p_\text{train}) - mc^2$$
which works out to
$$K_\text{ground} = \gamma K_\text{train} + (\gamma - 1) mc^2 + \gamma\beta c p_\text{train}$$
where $K$ is the relativistic kinetic energy and $p$ is the relativistic momentum.
If you wanted it in terms of energy alone:
$$K_\text{ground} = \gamma K_\text{train} + (\gamma - 1) mc^2 + \gamma\beta\sqrt{K_\text{train}^2 + 2 mc^2 K_\text{train}}$$
You might start to notice a similarity to the non-relativistic expression above ($K + \sqrt{2Km}V + \frac{1}{2}mV^2$), and indeed, if you plug in some approximations that are valid at low speeds ($\gamma \approx 1$, $\gamma - 1 \approx V^2/c^2$, $K_\text{train} \approx \frac{1}{2}mv^2 \ll mc^2$), you will recover exactly that expression.
A: Let's approach it from a different angle. Recall that kinetic energy, K, is defined as the mass-energy difference between the dynamic mass (in motion), m, and it's rest mass, m0, so that K = (m-m0)c^2. So the question simplifies to "Does the mass measured in one frame of reference equal that measured in another frame of reference."
Let's look at an example: Frame 1 is the earth. Frame 2 is a space ship with velocity, v, relative to earth. The spaceship is initially at rest on earth and contains a 1 kg test mass. The spaceship accelerates from rest to velocity, v. Both observer's on earth and inside the spaceship then measure the mass of the test mass. 
For the earth observer, the increase in the test mass follows Einstein's famous equation from Special Theory; m=m0/sqrt(1-v^2/c^2). The observer inside the spaceship measures
1) Inside the spaceship the observer knows he is accelerating away from earth. He can measure the acceleration, and can calculate he is traveling at velocity, v, relative to earth. Knowing this, he can calculate mass of the test mass using the ST relation. Same equation give the same result; the test mass increases by the same amount as that measured by earth observer above.
2) Trajectories of an accelerated particle are sometimes a function of its mass. For example within a cyclotron, the particle trajectory is a function of its mass. Furthermore, the trajectory is known to change as its mass increases with velocity.  Since there can be only one trajectory regardless of the frame of reference, all frames of reference must logically conclude the test mass has the same mass as that measured on earth.
Both frames of reference measure precisely the same increase in the test mass. That is to say, mass is invariant between two inertial frames of reference. If mass changes in one frame of reference, it likewise changes in the other frame of reference. Consequently, Kinetic energy being a function of mass increase, is not relative, rather is conserved.
