$\newcommand{\bs}{\boldsymbol}$First off the Standard Model (SM) is chrial, so left and right handed fermions are in different representations of the gauge group.
The rep of $SU(3)$ is determined by the color charge. Gluons are in the adjoint of $SU(3)$, which is the $\bs 8$ of $SU(3)$. Both left and right handed quarks are in the fundamental rep, which is a $\bs 3$ (or a $\bs{\bar{ 3}}$) (for example, the up quark field transforms as a $\bs 3$ of $SU(3)$; stated in more physics-y language, an up quark comes in three colors). All other fundamental fields in the SM are colorless, meaning they transform in the $\bs 1$ rep.
The rep of $SU(2)_L$ is determined by the transformation properties under the weak charge.The $W^\pm$ and $Z$ bosons transform under the adjoint, which is a $\bs 3$ rep. The left handed fermions transform. Within each generation, the different quarks mix and the different leptons mix. So $u_L$ and $d_L$ together form a $\bs 2$ rep of $SU(2)$, similarly $e_L$ and $\nu_{e,L}$ form a $\bs 2$ of $SU(2)$. The higgs forms a doublet (another $\bs 2$) of $SU(2)$, although the vev of the higgs breaks the $SU(2)_L\times U(1)$ down to the $U(1)$ of electromagnetism. The rest of the particles in the SM (in particular the gluons and the right handed fermions) are in the singlet of $SU(2)_L$.
Finally the rep of $U(1)$ is the hypercharge of the object. The hypercharge assignments are chosen ultimately to give the right electric charge, you can find the hypercharge assignments at Wikipedia.
So, putting this together, as an example a left handed up quark transforms as
\begin{equation}
{\rm left\ handed\ up\ quark}\sim( 3, 2,1/3)
\end{equation}
The full list of reps for each particle is a bit annoying to write out, although you can also construct this from the rules above.
Update
Based on the comment, let me give a quick example of how group theory translates into the Lagrangian.
Let's just do a simple case. Let's say we had a set of 2 complex scalar fields transforming in the $\bs 2$ of a global $SU(2)$ symmetry. The representation $\bs 2$ (not the $2$ in $SU(2)$) means that the scalar fields should be arranged in a 2d column vector, so instead of writing $\Phi$ and $\Psi$ to represent the fields, we write $\Phi_a$, where $a$ is an index. In other words, the two scalar fields become linked because of the existence of the symmetry.
Under an $SU(2)$ transformations,
\begin{equation}
\Phi_a \rightarrow U_{a}^{\ \ b} \Phi_b,
\end{equation}
where $U$ is unitary, $U U^\dagger = U^\dagger U = 1$.
Similarly, it is useful to define $\Phi^a = (\Phi_a)^\dagger$ (that is, $\Phi^a$ is a conjugate transpose of $\Phi_a$, so it is like a row vector). Then $\Phi^a$ transforms as
\begin{equation}
\Phi^a \rightarrow \Phi^b(U^\dagger)^{\ \ a}_{b}.
\end{equation}
Note that $\Phi^a \Phi_a$ is invariant under these $SU(2)$ transformations by unitarity.
Thus, you need to write lagrangians invariant under this transformation. This can by done by writing for example
\begin{equation}
\mathcal{L} = -\partial_\mu \Phi^a \partial^\mu \Phi_a - V(\Phi^a \Phi_a).
\end{equation}
If you want the theory to be renormalizable, then $V$ should only contain terms up to $O(\Phi^4)$ in 4 space-time dimensions.
The SM has lots of different reps, which at a pragmatic level shows up as lots of different kinds of indices on fields, each with their own transformation rule. The symmetries are also local, which involves coupling in the gauge bosons. I think if you want more detail you need to ask a separate question, the original question was about the representations present in the SM, the full Lagrangian of the SM is a completely different question.
I) The main point is that we usually only consider tensor products $V \otimes W$ of vector spaces $V$, $W$ (as opposed to general sets $V$, $W$). But groups (say $G$, $H$) are often not vector spaces. If we only consider tensor products of vector spaces, then the object $G \otimes H$ is nonsense, mathematically speaking.
With further assumptions on the groups $G$ and $H$, it is sometimes possible to define a tensor product $G \otimes H$ of groups, cf. my Phys.SE answer here and links therein.
II) If $V$ and $W$ are two vector spaces, then the tensor product $V \otimes W$ is again a vector space. Also the direct or Cartesian product $V\times W$ of vector spaces is isomorphic to the direct sum $V \oplus W$ of vector spaces, which is again a vector space.
In fact, if $V$ is a representation space for the group $G$, and $W$ is a representation space for the group $H$, then both the tensor product $V\otimes W$ and the direct sum $V\oplus W$ are representation spaces for the Cartesian product group $G\times H$.
(The direct sum representation space $V\oplus W\cong (V\otimes \mathbb{F}) \oplus(\mathbb{F}\otimes W)$ for the Cartesian product group $G\times H$ can be viewed as a direct sum of two $G\times H$ representation spaces, and is hence a composite concept. Recall that any group has a trivial representation.)
This interplay between the tensor product $V\otimes W$ and the Cartesian product $G\times H$ may persuade some authors into using the misleading notation $G\otimes H$ for the Cartesian product $G\times H$. Unfortunately, this often happens in physics and in category theory.
III) In contrast to groups, note that Lie algebras (say $\mathfrak{g}$, $\mathfrak{h}$) are always vector spaces, so tensor products $\mathfrak{g}\otimes\mathfrak{h}$ of Lie algebras do make sense. However due to exponentiation, it is typically the direct sum $\mathfrak{g}\oplus\mathfrak{h}$ of Lie algebras that is relevant. If $\exp:\mathfrak{g}\to G$ and $\exp:\mathfrak{h}\to H$ denote exponential maps, then $\exp:\mathfrak{g}\oplus\mathfrak{h}\to G\times H$.
Best Answer
Baez actually has another paper (with Huerta) that goes into more detail about this. In particular, Sec. 3.1 is where it's explained, along with some nice examples. The upshot is that the hypercharges of known particles work out just right so that the action of that generator is trivial. Specifically, we have
Because these are the only values for known fermions in the Standard Model, that generator does nothing. So basically, you can just take the full group modulo the subgroup generated by $(\alpha, \alpha^{-3}, \alpha^2)$ -- where $\alpha$ is a sixth root of unity.
There's also this paper by Saller, which goes into greater detail about the "central correlations" of the Standard Model's gauge group, but in a more technical presentation. Saller also goes into some detail in chapter 6.5.3 of his book.