He explains that the Higgs boson is a particular excitation mode of the Higgs field, but what is then the (general) 'ziggs' boson? Are Higgs and ziggs actually the same?
If there was only a massive $Z$ boson, and nothing else, then this 'ziggs' mechanism would be sufficient and the slightly more complicated 'Higgs' mechanism would not be required. I would look at the 'ziggs' as what the 'Higgs' would look like in this scenario.
The Z boson has been known experimentally for decades, what about its weakly hypercharged product? Does the Higgs boson actually have nothing to do with the Higgs phenomenon, being merely a consequence of the theory that was begging to be discovered experimentally?
I presume by 'hypercharged product' you mean the charged $W^+$ and $W^-$ bosons. These are not products of the $Z$ boson, instead they are all a part of the same family of particles. Well its actually a bit more complicated than that... let's start from the top. The Higgs mechanism 'starts' with four massless bosons ($B^0$,$W^0$,$W^1$,$W^2$), which after some interacting with the Higgs field, will produce three massive bosons ($Z$,$W^+$,$W^-$) and one massless boson (which is the photon, $\gamma$).
As there are four original bosons, without going into the heavier details, this means that the Higgs-field should be constructed from four components (or degrees of freedom). After interaction with the four components of the field, the four original bosons acquire a mass. However - as mentioned earlier - the photon is massless. In order to add this condition into our theory, we mix two of the original four bosons (the $B^0$ and $W^0$, which have no electric charge) to produce the the observed $Z$ and $\gamma$ bosons in such an exact way as to leave the photon massless. The repercussion of this is that we are left with one free component of our Higgs-field, remaining from our original four-component Higgs-field. This spare component, analogously to the 'ziggs' mechanism, manifests itself as the Higgs boson.
Normalization.
The action (and the Lagrangian) of a theory are defined up to a normalization factor (and other things, like total derivatives, but let us focus on multiplicative factors). There is no $\textit{correct}$ way to choose this normalization, you just have to stick with one and be coherent.
Suppose you have a real scalar field $\phi$: a Lagrangian for such a field will be
$$
-\frac12\partial_\mu\phi\partial^\mu\phi-V(\phi).
$$
When you derive w.r.t. $\partial_\mu\phi$, you have two fields to derive, and the factor of $\frac12$ gets cancelled in the motion equations:
$$
-\partial_\mu\partial^\mu\phi=-\frac{\partial V}{\partial \phi}.
$$
With this normalization, the LHS has no multiplicative factors, just the Laplacian of $\phi$.
Now let $\varphi$ be a complex scalar field. You interpret $\varphi$ and $\varphi^*$ to be two independent fields: when you vary with respect to one, you do not vary with respect to the other. A typical Lagrangian is now
$$
-\partial_\mu\varphi\partial^\mu\varphi^*-V(\varphi,\varphi^*),
$$
and motion equations are obtained by deriving w.r.t. $\varphi$ and $\varphi^*$:
$$
-\partial_\mu\partial^\mu\varphi^*=-\frac{\partial V}{\partial\varphi},\\
-\partial_\mu\partial^\mu\varphi=-\frac{\partial V}{\partial\varphi^*}.
$$
You do not have the factor $2$ in front of the kinetic term, as you do not need it to properly normalize the LHS of the equation. Now, try to plug
$$
\varphi=\frac{\phi_1+i\phi_2}{\sqrt 2}
$$
and analogous for $\varphi^*$ and see what happens.
About the sign: if you follow Srednicki, you're using $(-,+,+,+)$ as metric tensor. The minus in the front of the kinetic term is needed to have a definite kinetical energy. Remember that you can interpret the action as $S=\int(T-V)dt$: the terms with two time derivatives compose the kinetic energy, that enters with its sign in the action. The kinetic term must be a square of something, to have positive kinetic energy. To find it, we can write
$$
-\partial_\mu\phi\partial^\mu\phi=(\partial_0\phi)^2-(\partial_i\phi)^2
$$
Now, the terms in the parentheses are squares, so they are positive. The first parenthesis enters in the kinetical energy, and it's positive as we wanted it to. Notice that, if you had the metric $(+,-,-,-)$, you would have needed to change the sign of the first term in your Lagrangians: you always want a positive kinetic energy.
Best Answer
It is a non-issue, predicated on two conventions.
The historical convention defines it as $ Y_{\rm W} = 2(Q - T_3)$, as in the Gell-Mann—Nishijima formula of the strong interactions——for a conserved quantity. There, it was frequently used for strange particles, so the hypercharge could get to be 2, —2, etc... and a normalization like this one was warranted. In the weak interactions, thus, the weak hypercharge is defined as twice the average charge of a weak isomultiplet (where the average $T_3$ vanishes).
However, the more practical younger generation use $ Y_{\rm W} = (Q - T_3)$, instead, so the average charge of the isomultiplet, so, e.g., for right-handed fermions, weak isosinglets, the hypercharge is the charge, without daffy gratuitous 2s in the way. But it is only a matter of convention, and references such as the one you quote also specify the convention, as they should.