Why under Lorentz transformations the Higgs boson is a scalar field and under $SU(2)$ it is a doublet? I am a bit confused about this difference. My understanding is that when we build a $G$-bundle, where $G$ is a gauge group, we have a representation $\rho:G\to GL(V)$ that acts on the fibers of the $G$-bundle. Now if we want to act $SU(2),$ for example, on a scalar field $\phi$, we should use an one-dimensional representation since $\phi:M\to\mathbb{C}$, right? But how during this process the field acquires two components? I would say that $\phi:M\to \mathbb{C}$ and $\phi:M\to \mathbb{C}^2$ are sections of different bundles, so how can they be the same?
PS: I would appreciate an answer in terms of fiber bundles.
 A: I can't give an answer using fiber bundles, but I don't think it is important as the confusion is at a much simpler level. 
A field can be in different representations for different symmetry group. The Higgs field is in the trivial representation of the Poincarre group, that is, under Lorentz transformations, $\phi(x)\to \phi(\Lambda x)$, but in non-trivial representation of several internal symmetries (a doublet of SU(2) $\phi(x)\to U \phi(x)$,  in the non-trivial  representation of U(1), but in the trivial representation of SU(3) (associated to QCD)). Whether this symmetry is gauged or not does not matter here.
Edit: 
With the field described above (doublet of SU(2), trivial repr. of Poincarre and SU(3), non-trivial of U(1)), we find that we only need two complex fields $\phi(x)\equiv(\phi_1(x),\phi_2(x))$ (so that $\phi : M\to \mathbb{C}^2$). 
The transformations are thus ($a=1,2$):


*

*Lorentz : $\phi_a(x)\to \phi_a(\Lambda x)$, 

*SU(3) : $\phi_a(x)\to \phi_a(x)$,

*SU(2) : $\phi_a(x)\to U_{ab}\phi_b(x)$,

*U(1) : $\phi_a(x) \to e^{i \theta} \phi_a(x)$.

A: Let us consider an example and take the Weinberg-Salam Lagrangian:
$$
\mathscr{L} = i\bar{\psi}\gamma\cdot\partial\psi - m\bar{\psi}\psi
$$
and let us adapt it to the case describing electrons and neutrinos as
$$
\mathscr{L} = i\bar{\textrm{e}}_R\gamma\cdot\partial\textrm{e}_R + 
i\bar{\textrm{e}}_L\gamma\cdot\partial\textrm{e}_L + i\bar{\nu}_L\gamma\cdot\partial\nu_L
$$
where we have decomposed the spinor $\psi$ into its left (respectively right) projection and made use of the physical assumption that no right-helicity neutrinos have been found yet (thus they do not appear in the Lagrangian). All fields are massless, as mass will be generated through the Higgs mechanism and symmetry breaking. We assume, furthermore, that the internal group of transformations (if there is any) will map fields into fields with the same physical properties; therefore we claim
$$
L = \begin{pmatrix}\nu_L\\ \textrm{e}_L\end{pmatrix}\qquad R = \textrm{e}_R
$$
so that
$$
\mathscr{L} = i\bar{R}\gamma\cdot\partial R + i\bar{L}\gamma\cdot\partial L
$$
If we assume $R$ to transform under the one-dimensional representation of $SU(2)$ and L under the standard two-dimensional one, we can then claim the above Lagrangian in invariant under $SU(2)$ transformations. If so is the structure, then whatever additional field we introduce should at most be of similar form, hence the Higgs field can be written as
$$
\Phi = \begin{pmatrix}\phi^+(x)\\ \phi^0(x)\end{pmatrix}
$$
where the superscript $+,0$ is such because of the isospin (but we may have as well called them $\phi^{1,2}$). It is then clear, now, that in order to preserve the invariance of the initial structure, the new Higgs field must transform as
$$
\Phi' = \begin{pmatrix}{\phi^+}'\\ {\phi^0}'\end{pmatrix} = 
\begin{pmatrix}\ldots && \ldots \\ \ldots && \ldots\end{pmatrix}
\begin{pmatrix}{\phi^+}\\ {\phi^0}\end{pmatrix} = \Lambda(SU(2))\Phi
$$
where the big matrix is a two-dimensional representation of $SU(2)$.
Applying a local Lorentz transformation $\Gamma$ on the point $x$ upon which the components of the Higgs fields $\phi^{+,0}$ depend we have, though, 
$$
\Phi'(x') = \Phi'(\Gamma x) = \Phi(x)
$$
as per requirement that $\Phi$ is a scalar field in the Lorentz representation.
A: When we say scalar, spinor, vector, and so on, field, we mean which representation of the frame bundle the field belongs to. Or in index notation, which spacetime indices the field has: none, spinor, vector, and so on. We can combine this with internal symmetries which are $G$-bundles for some gauge group $G$, for example $SU(2)$. In indices this is some additional internal index. For example the gauge potential in QCD is usually written $A_{\mu a}$ where $\mu$ is the vector index and $a$ the color ($\operatorname{ad} SU(3)$) index. 
The way to do this is that if $E,F$ are vector bundles over $M$ then there exists a bundle $E \otimes F$ over $M$ such that the fiber is $e \otimes f$. The structure group of this product bundle is the product of the structure groups of $E,F$. Thus we can speak of things like an $SU(2)$ singlet scalar or an $SU(3)$ triplet spinor. In the former case $E$ is the trivial line bundle and $F$ the $SU(2)$ doublet bundle. [The proof of this theorem consists of writing the statement out in a local section and a checking that the transition maps work properly. For this what is needed is that $u \otimes v $ is smooth in both arguments using the usual notion of derivatives on finite-dimensional vector spaces. Thus the statement generalizes to functors like $\wedge,\oplus$. ]
