Consider the motion of a slow moving particle with worldline $x^a = x^a(t)$. We have
$$\frac{dx^0}{dt}=1, \frac{dx^1}{dt}=u_1,\frac{dx^2}{dt}=u_2,\frac{dx^3}{dt}=u_3$$
where $(u_1,u_2,u_3)$ is the velocity in the inertial coordinates on Minkowski space. In particular we have $\gamma(u) \sim 1$ and so we can identify the coordinate time $t$ with the proper time $\tau$ along the worldline, and so approximate the four-velocty by
$$(V^a)=(1,u_1,u_2,u_3)$$
The geodesic equation is
$$\frac{d^2x^a}{d\tau^2}+\Gamma_{bc}^a\frac{dx^b}{d\tau}\frac{dx^c}{d\tau}=0$$
Because we ignore products of the spatial components of the four-velocity with the Christoffel symbols, this is approximated by
$$\frac{d^2x^a}{d\tau^2}+\Gamma_{00}^a=0$$
We also ignore terms involving time $(x^0)$ derivatives of the metric components.
1) Where has the $\gamma(u) \sim 1$ come from?
2) a) are we ignoring products due to being slow moving?
b) I still cant see how the last equation has been derived
3) We do we only ignore components of the time derivative? Why not the others?
A short answner:
For your first question, here a hint: consider that $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$ where $\beta = v/c$, approximate $\gamma$ when $\ v << c$ (Slow moving particle).
for the second and third question's:
yes, $\frac{dx^i}{d\tau}\frac{dx^j}{d\tau}=\gamma^2 u^i u^j$, $\gamma^2 \approx 1$, then $\frac{dx^i}{d\tau}\frac{dx^j}{d\tau}=u^i u^j$, if the particle's velocity is sufficiently small, the product of these two vectors is negligible, then we consider only the "time component" i.e. $\Gamma_{00}^a\frac{dx^0}{d\tau}\frac{dx^0}{d\tau}$, but $\frac{dx^0}{dt}=1$, then $\frac{d^2x^a}{d\tau^2}+\Gamma_{00}^a=0$.
about the third question, when the spacetime is static i.e. time-independent, all time-derivatives of the metric vanishes.
*Obs: the $ \gamma$ factor that i mentioned is called Lorentz factor, here a Wikipedia's article about this.