Runge-Kutta method and step doubling
Answer to Part 1
The fractional coefficients in RK4 sum to 1. This is in part to satisfy what is known as the order conditions of the integrator. Basically, we want RK4 to be a 4th order method. For this to be true, it has to satisfy a special relationship using the coefficients of its Butcher Tableau (see at the top here), namely
$$\mathbf{b}^T A^kC^{l-1}\mathbf{1} = \frac{(l-1)!}{(l+k)!}$$
where $A$ is the coefficients in the main part of the tableau, $\mathbf{b}$ is a vector of the $b$ coefficients (in RK4, this is 1/6, 1/3, 1/3, 1/6), and $C$ is a diagonal matrix formed by the $C$ coefficients.
We need to craft these coefficients to get $0$-stability. It is provable that if the method is $0$-stable, it will converge to the order of accuracy -- in this case, order 4, or $h^4$.
The reason that we carry $k_1$, $k_2$, etc. through to the final solution is because $k_4$ is not simply a sum of the previous $k$'s. Each $k$ can be thought of as a projection of a finite difference. The RK method essentially averages these projections in a non-standard way. If you do out the math formally, you find that with these weights, you get optimal accuracy for your number of function evaluations.
For the first part, the Wikipedia article http://en.m.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods has a detailed discussion of the different members of the family along with a derivation of RK-4. The key idea is that you can do whatever you want in terms of writing down the form of your approximation; all that matters is that doing a Taylor expansion makes all the error terms up to the order of approximation vanish. This is how the coefficients are chosen.
Then in the next part, $\phi$ is the the expression for the error made in the step, without the $(2h)^5$ term. It should be thought of as an expression in terms of the derivatives of $y$ at $x$, but its exact form isn't important, just its independence from $h$. It's the same in both lines to first order in $h$ so the errors are hidden away in the correction term. Note that fourth order means the error term is $h^5$ per step which it is throughout. You then use two fourth order approximations to get a better one by calculating the error approximately. This is reasonable since you've used more data.
As to the $1/15$ factor, that simply arises because $(2h)^5-2(h)^5=(16-1)h^5=15h^5$ and if you make only a step of size $h$ you get a factor of $h^5$. Try carefully defining and writing down the different things you work out, then you can see why you take the difference in this way.