Whenever we say Higgs boson we actually mean "the quantum of the Higgs field, post SSB$^\dagger$". In the same way the "photon" is usually associated with the quantum of the post-SSB $A^\mu$ field, and not the pre-SSB (mixed) $B^\mu$.
The particles ( the quanta of the fields) are just excitations of the fields. The Higgs field operator $\hat{\phi}_H(\mathbf{x})$ will create one Higgs boson at position $\mathbf{x}$. The field itself, however, is the more imporant entity, and it permeates the whole of space.
If the field has $>1$ degrees of freedom, its quanta have $>1$ internal configurations.
The photon field $A^\mu$ has $2$ degrees of freedom: this corresponds to two mutually orthogonally polarised photons.
The $4$ degrees of freedom of the pre-SSB Higgs field can be thought as "polarisations" of the pre-SSB Higgs boson. Though this has imaginary mass so a physical interpretation is somewhat challenging (but let me know if you come up with one).
I think they key problem with you is the Higgs field before and after SSB.
Before SSB:
The Higgs field is a $SU(2)$ complex doublet, as you said, meaning it is a $2$-component quantity, each entry being complex:
$$ \phi_H = \left ( \begin{array}{c} \phi^+ \\ \phi^0 \end{array} \right ) = \left ( \begin{array}{c} \phi^+_1 + i\phi^+_2 \\ \phi^0_1 + i \phi^0_2 \end{array} \right ),$$
with $\phi^+, \phi^0 \in \mathbb{C}$ and $\phi^+_1, \phi^+_2, \phi^0_1, \phi^0_2 \in \mathbb{R}$.
So there $4$ independent parameters, or "degrees of freedom" (d.o.f.).
Each can be considered as a scalar field, yes -- scalar fields carry one degree of freedom each, so the maths checks out.
The Lagrangian density $\mathcal{L}$ has an $SU(2)\times U(1)$ symmetry, which has $4$ generators (different $4$ from before).
Upon requiring this symmetry to be local, you introduce $4$ gauge fields $W^1, W^2, W^3, B$.
These have to be massless since bare mass terms for gauge bosons are not allowed in the Standard model, as they would violate gauge symmetry. A massless gauge field has $2$ degrees of freedom (two polarisations, think of a photon).
So: $4$ massless gauge fields ($8$ d.o.f.) $+$ $1$ complex Higgs doublet ($4$ d.o.f.) = $12$ d.o.f.
Symmetry: $SU(2)_L \times U(1)_Y$.
After SSB:
The Higgs mechanism happens.
It breaks $SU(2)_L \times U(1)_Y$ into $U(1)_{\mathrm{em}}$. The number of broken generators are $4-1$ = $3$.
These would-be Goldstone bosons are "eaten up" by $3$ of the massless gauge fields, which now become massive. The Higgs mechanism also mixes these fields around, so now they are labelled $W^+, W^-, Z^0$ and $A$, the latter being the only one retaining its massless status (the photon field).
Massive gauge fields have $3$ degrees of freedom, think of an atom with spin $1$: it has projections $-1, 0, 1$.
In the process, the Higgs complex doublet reduces to a single real scalar field, what is usually referred as "the Higgs boson".
So: $3$ massive gauge fields ($9$ d.o.f.) + $1$ massless gauge field ($2$ d.o.f.) + $1$ real scalar field ($1$ d.o.f.) = $12$ d.o.f.
The number of degree of freedom of the $SU(2)_L \times U(1)_Y$ theory (electro-weak) remains unchanged. All that happens is a "reshuffling" around.
The Higgs boson pre-SSB, i.e. the quantum of the pre-SSB Higgs field, is a single particle carrying $4$ degrees of freedom ("polarisations"). Though it has imaginary mass so physical interpretation is challenging.
The Higgs boson post-SSB, i.e. the quantum of the post-SSB Higgs field, is a single particle carrying $1$ degree of freedom. This has a real mass term, and is what we associate with the experimentally observed particle.
$\dagger$ = Spontaneous Symmetry Breaking.
