$x$ is a position vector here.
$$ f(x) = -\frac{x}{|x|^3} $$Newton equation
$$ \frac{d^2 x}{dt^2} = f(x) = -\frac{x}{|x|^3} $$Define an intermediate variable to make it first-order differetial equation $ v \equiv \frac{dx}{dt} $
$$ \frac{dv}{dt} = -\frac{x}{|x|^3} \\ \frac{dx}{dt} = v $$Make the differential deltas
$$ \Delta v = \frac{x}{|x|^3} \Delta t \\ \Delta x = v \Delta t $$Formal definition:
$$ x_{n+1} = x_n + hf'(t_n, x_n) $$In our case:
$$ v_{n+1} = v_n + h \frac{x}{|x|^3} \\ x_{t+1} = x_n + h v(t) $$Formal definition:
\begin{align*} k_1 &= h f'(t_n, x_n) \\ k_2 &= h f'(t_n + \frac 12 h, x_n + \frac 12 k_1) \\ x_{n+1} &= x_n + k_2 \end{align*}In our case:
\begin{align*} \Delta v_1 &= h f(x) \\ \Delta x_1 &= h v(t) \\ \Delta v_2 &= h f(x + \Delta x_1 / 2) \\ \Delta x_2 &= h (v + \Delta v_1 / 2) \\ v(t+h) &= v(t) + \Delta v_2 \\ x(t+h) &= x(t) + \Delta x_2 \end{align*}Formal definition:
\begin{align*} k_1 &= h f'(t_n, x_n) \\ k_2 &= h f'(t_n + \frac 12 h, x_n + \frac 12 k_1) \\ k_3 &= h f'(t_n + \frac 12 h, x_n + \frac 12 k_2) \\ k_4 &= h f'(t_n + h, x_n + k_3) \\ x_{n+1} &= x_n + \frac{k_1}{6} + \frac{k_2}{3} + \frac{k_3}{3} + \frac{k_4}{6} \end{align*}In our case: ?