Describing abelian symmetry breaking in his book on gauge theories, after favouring a vacuum (whose expectation value is $v$) from the symmetric continuum, Quigg parametrize the complex scalar as
$$ \phi=e^{i\frac{\zeta}{v}}\frac{v+\eta}{\sqrt{2}} \tag{1} $$
He then proceeds to use the gauge freedom to eliminate the Goldstone (the gauge boson is irrelevant to my doubt here) with a transformation whose gauge parameter is $\alpha=-\frac{\zeta}{v}$.
My problem is the non-abelian case. Suppose a $SU(2)$ model in which the abelian subgroups corresponding to the first two generators $T_1$ and $T_2$ were broken. Then, analogously to (1), what is done is writing the field as
$$ \phi=e^{\frac{i}{v}(\zeta_1T_1+\zeta_2T_2)} \left( \begin{matrix} 0 \\ 0 \\ v +\eta\end{matrix} \right) \tag{2} $$
after which we go to the U-gauge by means of a gauge trasformation with parameter $\vec{\zeta}=-\frac{1}{v}(\zeta_1, \zeta_2, 0)$.
I do not understand equation (2). It is clear that what we do after shifting the vacuum component around its expectation value is to write the field as the resulting vector acted on by the exponential of a arbitrary linear combination of the broken generators, but I don't know why that is the case. At first, equation (1) looked to me as just writing the field in a polar form, by convenience, since it is absolutely and obviously general. I would then expect that (2) is again just a general way of expressing a 3-component complex arbitrary vector, but I can't see why this is a general fact. That is, the third component is taken care of by the 'arbitrary' $\eta$, but I don't know if the first two are taken care of by the exponential.
Of course, my way of thinking and of interpreting (1) might just be plain wrong. So, why does the exponential in (2) carry only the broken generators?
(Edit) Example: In his book, Frampton deals with a $SO(N)$ model: breaking the (associated subspace of the) first $N-1$ generators, he chooses the vacuum
$$ \phi_0=\left(\begin{matrix} 0 \\ 0 \\ ... \\ v+\eta \end{matrix}\right) $$
and using the explicit generators $(L_{ij})_{kl}=-i(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})$ he shows that we can parametrize
$$ \phi=\exp\left({\frac{i}{v}\sum_{i=1}^{N-1}\xi_{i}T_i}\right)\left(\begin{matrix} 0 \\ 0 \\ ... \\ v+\eta \end{matrix}\right) = \left(\begin{matrix} \xi_1 \\ ... \\ \xi_{N-1} \\ v+\eta \end{matrix}\right) $$
where $T_i \equiv L_{iN}$. After this, he goes to the U-gauge by the standard procedure. The point is that in this example what happens is exactly what I was under the impression of being the case: the (imaginary argument) exponential of a linear combination of the broken generators just happens to, when acting on the vacuum, generate a general (modulo the fact that the $\xi_i$ look to be phases) vector in the $N-1$ first entries. How general is this and why?
(Edit2): A long time ago I had accepted the answer below (which is very thoughtful and greatly appreciated), but rereading it now I realize I have actually never understood it. I will try to identify some specific confusions of mine regarding it (of course, you can also just ignore this part of the question and answer its core, above, if you think the answer below is inadequate).
(i) Take the passage "Fields transforming under G furnish a representation and they are elements of G. Then, you can make the identification ϕ=g and apply the decomposition Eq. (1).": How are fields transforming under some representation identifiable with the elements of the representation itself (to the point of formally equating to the transformations' form)?
(ii) Equation (1) seems to be a quite general and powerful result. What is its validity (meaning, are there other conditions than the ones stated needed for it to work) and how would one prove it?
TL;DR Question: To state briefly the question: the operational beauty with the higgs formalism is that it is easy to go to the unitary gauge where the Goldstones manifestly decouple by means of a gauge transformation, but how and why can one always parametrize a general expansion of the field around the favored vacuum as something like
$$\text{exp}\left\{i\sum_{\text{broken}} \xi_i T^i\right\}\left(\begin{matrix} ... \\ v+H \end{matrix}\right),$$
where $T_i$ are the generators and $H$ is the 'Higgs', so that a gauge transformation may cancel the exponential and the spectrum is obvious.
