You seem want to introduce gauge invariance into a theory that doesn't appear to have the global symmetry need in the first place. One way to think of gauge invariance is that you 'gauge' the global symmetry, then you just change your derivative terms to covariant derivatives like you mentioned. In other words, we can only concern ourselves with the global symmetry for now and gauge it at the very end if we wish. Now, at the very least in the term you wrote down
$$\phi \bar{\psi} \psi$$
it's not clear how the indices are contracted. Do the $\phi$ and $\psi$ have indices? For example I could make the $\psi$ transform in under $SU(2)$ by introducing a second copy of the $\psi$ and sum over them:
$$ \phi \bar{\psi^a} \psi^a $$
but now I can't make the $\phi $ transform because there isn't anything left to contract the $\phi$ index with. That is,
$$ \phi^b \bar{\psi^a} \psi^a $$
isn't a singlet (a singlet doesn't transform under the symmetry ) so it doesn't make any sense as a term in your lagrangian. Or you could introduce a second spinor that is a singlet under the global symmetry so you have something to contract your scalar indices with:
$$ \phi^a \bar{\psi^a} \eta $$
Finally, if you want a spinor to transform as a triplet, or in the adjoint of $SU(2)$ you could introduce the generator of SU(2) and contract indices as follows:
$$ \phi^a (t^a)^{bc} \bar{\psi^b} \psi^c = \phi^a \bar{\psi} t^a \psi$$
where the $t^a$ is in the fundamental (or doublet) representation so that we can properly trade the adjoint index for two doublet indices and get an overall singlet for the lagrangian.
Now, as for the kinetic terms, like you mentioned, if you want to introduce gauge symmetry trade the regular derivative for a covariant derivative.
$$ \partial_\mu \phi^a \rightarrow D_\mu \phi^a =\partial_\mu \phi^a + i {A_\mu}^b (t^b)^{ac}\phi^c $$
where the $t^a$ is in whatever rep the $\phi^a$ transforms in. The same holds for the fermion fields.
There is one caveat to all this however - this prescription of naively gauging a global symmetry that I have outlined breaks down if the global symmetry is 'anomalous'. That is, quantum mechanical effects break the naive, classical global symmetry. I'm not going to get into what that is, but keep it in the back of your mind for the time being and read about it when you have a chance.
I have a feeling you might want more info than this, but I'll stop here for now and if you edit I will clarify/ add.
EDIT: In retrospect this seems to work more easily for $SU(2)$ reps easier than other groups since for $SU(2)$ the adjoint rep is the same as the triplet rep so I can trade triplet rep indices for doublet indices using the generators $(t^a)^{ij}$. I am not sure if you can do these sort of things for groups in general.
Best Answer
$\newcommand{\bs}{\boldsymbol}$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.