There are various layers here.
First, the Higgs boson is a EW doublet with a certain hypercharge $Y$. Defining as usual the electric charge as $Q=T^3+Y$ you see that $\phi^0$ is indeed neutral for the choice $Y=1/2$. But you see also that for the very same choice, for which a vev of $\phi_3$ doesn't break the charge $Q$, the other component $\phi^+$ must be charged. In fact, must have charge exactly $+1$, as given by $Q=T^3+Y=\mathrm{diag}(1,0)$.
Second, even without knowing what the charges are, as long as the charge $Q$ is not broken, the 3 polarizations for every $W$ must have the same electric charge.
Finally, you can see that $\phi_{1,2,4}$ are indeed ``eaten'' by changing variables and using instead
$$\phi=e^{i\pi(x)/v \hat{T}}\left(\begin{array}{cc}0\\ \frac{v+h}{\sqrt{2}}\end{array}\right)$$
where $\hat{T}$ are the broken generators. Expanding in the $\pi(x)$ field the expression above for $\phi(x)$ you see that they are, at leading order, the $\phi_{1,2,4}$ (that is, the $\pi(x)$ generate on the vacuum states with the same quantum numbers as the $\phi_{1,2,4}$ do).
By doing a gauge transformation that schematically looks like
$$W_\mu\rightarrow e^{i\pi(x)/v \hat{T}}W_\mu e^{-i\pi(x)/v \hat{T}}-e^{i\pi(x)/v \hat{T}}\partial_\mu e^{-i\pi(x)/v \hat{T}}$$
(hoping that I dind't mess up the signs and other factors) you can remove the $\pi(x)$ fields, that is removing the $\phi_{1,2,4}$. As you can see from the gauge transformation, they give rise to non-vanishing longitudinal terms, such as $\partial_\mu\pi(x)/v$, in the expression for the $W$. In other words, a massive $W$ emerges via the so-called Higgs mechanism.
The hypercharge of a doublet cannot be "deduced". When one builds a gauge theory, the first step is to define the particle content of your theory and to postulate the representation of all particle multiplets. In particular, if the gauge group is abelian, then we have to assign numbers usually called charges.
So, I reformulate your question: Why do we choose the hypercharge of the Higgs doublet to be $Y=1$? The idea is that we want to respect the Gell-Mann–Nishijima formula $Y=2(Q-T_3)$. Since we know the charges and the values of $T_3=\pm 1/2$ for the doublet, it is straightforward to find that $Y=1$.
There are several ways to see why the formula $Y=2(Q-T_3)$ must be respected:
- One possibility is to compute the commutator of the generators $Q$ and $T_i$. You will notice that $[Q,T_i]=[T_3,T_i]\neq0$, but $[Q-T_3,T_i]=0$. This means that the global symmetries $SU(2)$ and $U(1)_{\rm em}$ cannot be simultaneously satisfied, but we can define the hypercharge $Y$ as new Abelian symmetry.
- Other possibility is to try to assign Abelian charges for the
fermionic multiplets of the Standard Model (knowing that the charge
must be the same for all members of a multiplet). You will conclude that $Y=2(Q-T_3)$ is the only possible choice up to a multiplicative
constant.
Best Answer
The two field doublets you are writing down involve the same four $\phi_i$s, and they constitute different arrangements of them, so they make no difference in the amplitudes involved and their physics implications. (The physical Higgs particle is linked to $\phi_3-v$.)
They are all "present" in our universe, as would any other rewriting of them, such as $\tilde \Phi$, etc... As long as the interactions and couplings of each are specified clearly, and each is fit into the SM we know, there should be no issue. Recall both $\Phi$ and $\Phi^*$ are present in the real covariant kinetic term in the SM Lagrangian written in the practical notation.