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].  

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 The Newtonian equation is not strictly correct and is superseded by the relativistic equation, but for velocities encountered in every day mechanics the Newtonian equation is a reasonable and convenient approximation.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

$$ \frac{x}{\sqrt{1-x} } = 2 \left(\frac{1}{\sqrt{1-x}} -1 \right)  =  \frac{2 (1 - \sqrt{1-x})}{\sqrt{1-x}}  $$

$$ {x}  = 2 (1 - \sqrt{1-x})  $$

$$ 2 \sqrt{1-x}  = 2-x $$

$$ 4 (1-x)  = (2-x)^2 = 4 -4x + x^2 $$

$$ 4 - 4 x - 4 + 4 x -x^2 = 0$$

$$  -x^2 = 0$$

$$  x = 0$$

  [1]: https://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy