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-Furthermore, it turns out that for a system with a minimum possible energy, only terms up to the first time derivative are possible. I refer you to an excellent previous question that discusses this.

Edit: reading your comments to your question, I should emphasize that this is not a definition of energy, although this Lagrangian formalism turns out to be useful for that too. Take a look at this question for some interesting discussion about the best way to define energy.

-Furthermore, it turns out that for a system with a minimum possible energy, only terms up to the first time derivative are possible. I refer you to an excellent previous question that discusses this.

Edit: reading your comments to your question, I should emphasize that this is not a definition of energy, although this Lagrangian formalism turns out to be useful for that too. Take a look at this question for some interesting discussion about the best way to define energy.

As you can see, this exact form has the interesting aspect that the mass energy and kinetic energy end up having a combined form, so it is somewhat natural to consider the mass energy as the constant part of the kinetic energy expression. But using the Taylor expanded expression one could justify grouping it with kinetic energy, potential energy, or as a separate category, so if you want to consider it as a third type of energy you can do that too.

This analysis was for a single particle, but field theories also show a similar division of the energy into terms involving derivatives of the field value and those involving the field value directly.

As you can see, this exact form has the interesting aspect that the mass energy and kinetic energy end up having a combined expression, so it is somewhat natural to consider the mass energy as the constant part of the kinetic energy expression. But using the Taylor expanded expression one could justify grouping it with kinetic energy, potential energy, or as a separate category, so if you want to consider it as a third type of energy you can do that too.

This analysis was for a single particle, but field theories also show a similar division of the energy into terms involving derivatives of the field value and those involving the field value directly, which can be considered as generalizations of the kinetic/potential energy division.

-Now all the terms are like $\dot{q}^n q^m$, for $n,m$ as non-negative integers. However, many of these terms have no physical effect, because they can be made to vanish by a suitable integration by parts. This is true because it is actually the integral of the Lagrangian, $\int L dt$, that is physically significant (this is the action, which obeys the Principle of Least Action)*. Once this criterion is applied, only terms with $n=0$ or $n\leq 2,m=0$ remain.

$L=\alpha\dot{q}^2+\beta\dot{q}-V(q)$

Here $-V(q)$ is a general function of only position, which I've defined with a minus sign for reasons that will become clear.

The prescription to find the energy of the particle from this (or, more precisely, the so-called Hamiltonian), is:

So just plug this in:

$H=(2\alpha \dot{q} +\beta )\dot{q}-(\alpha\dot{q}^2+\beta\dot{q}-V(q))$

$=\alpha \dot{q}^2+V(q)$

We end up with just a function of the velocity squared and the position. These can now be defined as the kinetic and potential energy. In particular, in this nonrelativistic setting $\alpha=m/2$ by definition.

-Now all the terms are like $\dot{q}^n q^m$, for $n,m$ as non-negative integers. However, many of these terms have no physical effect, because they can be made to vanish by a suitable integration by parts. This is true because it is actually the integral of the Lagrangian, $\int L dt$, that is physically significant (this is the action, which obeys the Principle of Least Action)*. Once this criterion is applied, only terms with $n=0$ or $m=0$ remain. (Note that this statement has been corrected)

$L=f(\dot{q})-V(q)+C$

Here $f(\dot{q})$ and $-V(q)$ are general functions of only velocity and position. I've also added explicitly a constant term, $C$, although this could have also been absorbed into the form for either $f$ or $V$. Now I will Taylor expand $f$:

$L=(\alpha \dot{q} + \beta \dot{q}^2+\gamma \dot{q}^3+\dots)-V(q)+C$

The prescription to find the energy of the particle from this (or, more precisely, the so-called Hamiltonian), is:

So just plug this and simplify for:

$H=(\beta \dot{q}^2+2\gamma \dot{q}^3+\dots)+V(q)-C$

We end up with just a function of the velocity (with lowest term $\sim \dot{q}^2$) and the position. These can now be defined as the kinetic and potential energy. The constant term, again, could be grouped with either.

At sufficiently low velocity, we should expect that only the lowest order term in the kinetic energy is relevant. Then we have an energy like:

$H=\beta \dot{q}^2+V(q)-C$

Defining $\beta=m/2$, this recovers the normal non-relativistic form of the energy of a particle. $C$ has no effects on the dynamics in this limit, so it does not matter what value it is given.

However, for a relativistic particle the higher-order kinetic energy terms do matter, and one ends up with an energy like:

$$H=(\frac{1}{2}m\dot{q}^2+\frac{3}{8}m\frac{\dot{q}^4}{c^2}+\dots)+V(q)+mc^2$$ $$=\frac{mc^2}{\sqrt{1-\frac{\dot{q}^2}{c^2}}}+V(q)$$

As you can see, this exact form has the interesting aspect that the mass energy and kinetic energy end up having a combined form, so it is somewhat natural to consider the mass energy as the constant part of the kinetic energy expression. But using the Taylor expanded expression one could justify grouping it with kinetic energy, potential energy, or as a separate category, so if you want to consider it as a third type of energy you can do that too.

This analysis was for a single particle, but field theories also show a similar division of the energy into terms involving derivatives of the field value and those involving the field value directly.