I know that the answer is $\mu_p \times \mathbb{Z}/p\mathbb{Z}$ if $E$ is ordinary, and $\alpha_p$ if $E$ is supersingular, where $\mu_p$ and $\alpha_p$ are the kernels of Frobenius on $\mathbb{G}_m$ and $\mathbb{G}_a$ respectively. But why is it this true?

Suppose $E'$ is a lift of $E$ to characteristic 0. Then $E'[p] = (\mathbb{Z}/p\mathbb{Z})^2$. If $E$ is ordinary then we have $E[p] =\mathbb{Z}/p\mathbb{Z}$, and one way to reconcile these two facts is to have $E[p] = \mu_p \times \mathbb{Z}/p\mathbb{Z}$, since the group of closed points of $\mu_p$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$ in characteristic 0, whereas in characteristic p, it is just a single (nonreduced) point. I'm not sure why this is the only possibility though--I think it has to do with the height of the formal group, but I just can't nut out the details. In Katz-Mazur "Arithmeticic moduli of elliptic curves" (proof of theorem 2.9.3) they say that "any p-divisible group over an algebraically closed field is the product of a p-divisible commutative formal Lie group and finite number of copies of $\mathbb{Q}_p/\mathbb{Z}_p$," but I don't see why this is true.

For the supersingular case, I'm even hazier. Is $\alpha_p$ the unique one-paramater formal group of height 2, and if so, how can you see this? For an affine scheme Spec(R), we have $\alpha_p(R) = \mathrm{Spec}(R[x]/(x^p))$. Is it true that in characteristic 0 we have (for example) $\alpha_p(\mathbb{C}_p) = (\mathbb{Z}/p\mathbb{Z})^2$?

Sorry if this is a mess, i'm really confused, and I haven't been able to find a sufficiently dumbed down explanation of this stuff anywhere.