In special relativity, the total energy of an object is given by; $$E = \sqrt{ (pc)^2 + (m_0 c^2)^2} = \sqrt{(\gamma m_0 v c)^2 +(m_0 c^2)^2}$$ where $p$ is the relativistic momentum $(\gamma m v)$.
The kinetic energy is the total energy minus the rest energy, so:
$$E_K = E - m_0 c^2 = \sqrt{(\gamma m_0 v c)^2 +(m_0 c^2)^2} - m_0 c^2$$
If you rearrange this equation you get : $$ E_K^2 + 2 E_K m_0 c^2 - (\gamma m_0 v c)^2 =0 \ .$$ This is a quadratic equation and the positive root is $$E_k = m_0 c^2 (\gamma -1) $$ in agreement with your first equation and as documented by [Wikipedia][1]as documented by Wikipedia.
If you take the Taylor expansion of this expression, the first three terms are: $$0 + \frac 1 2 m_0 v^2 + \frac 3 8 \frac{m_0 v^4} {c^2} $$
When v is small compared to c the terms with powers of v greater than 2 become negligible and the equation approximates to the Newtonian expectation $ 1/2 m v^2$. The Newtonian equation is not strictly correct and is superseded by the relativistic equation, but for velocities encountered in everyday mechanics the Newtonian equation is a reasonable and convenient approximation.
Your second equation $\gamma (1/2 \ m_0 v^2)$ is not equal to either the relativistic or the Newtonian equation for kinetic energy. For example, if we assume $$1/2 \gamma m v^2 = m_0 c^2 (\gamma -1) $$ this equivalence eventually reduces to $$\frac{v^2}{c^2} = 0$$ which can only be true if v=0.
How come it reduces to $v^2/c^2=0$ ?
$$1/2 \gamma \ m v^2 = m_0 c^2 (\gamma -1) $$
$$ \gamma \ v^2/c^2 = 2 (\gamma -1) $$
Substitute x for v^2/c^2
$$ \frac{x}{\sqrt{1-x} } = 2 \left(\frac{1}{\sqrt{1-x}} -1 \right) = \frac{2 (1 - \sqrt{1-x})}{\sqrt{1-x}} $$$$ \frac{x}{\sqrt{1-x} } = \frac{2 (1 - \sqrt{1-x})}{\sqrt{1-x}} $$
$$ {x} = 2 (1 - \sqrt{1-x}) $$
$$ 2 \sqrt{1-x} = 2-x $$
$$ 4 (1-x) = (2-x)^2 $$
$$ -x^2 = 0$$
$$ x = 0$$
[1]: https://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy
