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 The Newtonian equation is not strictly correct and is superseded by the relativistic equation, but for velocities encountered in every day mechanics the Newtonian equation is a reasonable and convenient approximation.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