The scalar potential of your theory is
$$V(\phi) = \lambda (\phi^a \phi^a - v^2)^2,$$
where I suspect you meant to take the square as I've written here. This potential is minimized when $\sqrt{\phi^a \phi^a} = v$. Think of $\phi=\frac{1}{2} \phi^a \sigma^a$ as a vector with components $\phi^a$ in a 3-dimensional vector space with basis vectors $\sigma_a/2$. The equation $\sqrt{\phi^a \phi^a}=v$ says that the norm of this vector is $v$ at the minima of the potential. The minimal locus of the potential is therefore a 2-sphere of radius $v$ in this 3d space, consisting of vectors whose norm is fixed to $v$.
Choose at random one of these minimal field configurations, say $\phi_0=\frac{1}{2} v \sigma^3$ (the "north pole" of the 2-sphere, if you like). Consider how a gauge transformation acts on this field configuration. Since the scalar transforms in the adjoint representation of the gauge group, a gauge transformation $e^{iT} \in \mathrm{SU}(2)$ acts on $\phi_0$ as $\phi_0 \mapsto e^{i T} \phi_0 e^{-iT}$, or, infinitesimally, $\delta \phi_0 = i [T,\phi_0]$. Here, $T = \frac{1}{2} T^a \sigma^a$ is an arbitrary element of $\mathrm{su}(2)$, the Lie algebra of $\mathrm{SU}(2)$. Our field configuration $\phi_0 = \frac{1}{2} v \sigma^3$ is therefore invariant under this gauge transformation when $T \propto \sigma_3$, while $\phi_0$ is not invariant if $T$ has support along $\sigma^1$ or $\sigma^2$. We find that a single generator of $\mathrm{su(2)}$ (which generates a $\mathrm{U(1)}$ subgroup of $\mathrm{SU(2)}$) leaves $\phi_0$ invariant, while the other two generators act non-trivially.
In such a situation, we say the $\mathrm{SU(2)}$ gauge symmetry has been Higgsed to a $\mathrm{U(1)}$ subgroup. To determine the spectrum of fields in the Higgsed theory, perform a field redefinition $\phi' = \phi- \phi_0$, such that the potential is now minimized for $\phi' = 0$. If you replace $\phi$ by $\phi'+\phi_0$ in your Lagrangian and expand it, you will find that the gauge fields $A_1$ and $A_2$ have become massive (in unitary gauge), while $A_3$ remains massless. You can immediately determine the gauge field masses by simply replacing $\phi$ by $\phi_0$ in the scalar kinetic term $(D_\mu \phi^a) (D^\mu \phi^a)$:
\begin{align}(D_\mu \phi_0^a)(D^\mu \phi_0^a) =& (g \epsilon^{abc} A_\mu^b v \delta^{c,3})(g \epsilon^{ade} A_\mu^d v \delta^{e,3})\\
=& g^2 v^2 \epsilon^{ab3} \epsilon^{ad3} A_\mu^b A_\mu^d\\
=&g^2 v^2 (A_\mu^1 A_\mu^1+A_\mu^2A_\mu^2).
\end{align}
Thus, $A_1$ and $A_2$ have each acquired a mass of order $gv$. They are often traded for the "complex gauge fields" $W_\pm = A_1 \pm i A_2$ because it is $W_\pm$ which appears in cubic interactions with matter. I will leave the explicit expansion of the Lagrangian for you to work out.
That should resolve your first two questions. Your third question asks what will happen if you take the scalar to live in a different representation. The analysis proceeds in the same way, so let me summarize the procedure for an arbitrary gauge group and representation. One begins with a scalar potential $V(\phi)$ and determines the fields configurations which minimize it, $\mathcal{M}_0 = \left \{\phi_0: V'(\phi_0) = 0, V''(\phi_0) > 0 \right\}$. Suppose the action is invariant under a symmetry group $G$, and that $\phi$ belongs to a linear representation $R$ of $G$. In other words, $\phi$ transforms as $\phi \mapsto R(g) \phi$, where $R(g)$ is the matrix representation of $g$. For an infinitesimal transformation (when $G$ is continuous), $\delta \phi = i T \phi$, where $R(g)=e^{iT}$.
Since the action is invariant under $G$, $V(\phi_0) = V(R(g) \phi_0)$ for any $g \in G$.
Thus $G$ maps $\mathcal{M}_0$ to itself. $R(g)$ need not leave $\phi_0$ invariant, however. In general, only a subgroup $H \subset G$ will leave $\phi_0$ invariant. That is, among the list of generators $\{T_a\}$ of $G$, some subset (the "unbroken" generators) will leave $\phi_0$ invariant, $\delta \phi_0 = i T_a \phi = 0$, while the remainder will not (the "broken" generators). The unbroken generators generate the subgroup $H$, while the broken generators correspond to the coset $G/H$.
If $G$ is a global symmetry, we say that it has been spontaneously broken to the subgroup $H$. If $G$ is a gauge symmetry, we say that it has been Higgsed to $H$. The gauge fields along the generators of $H$ remain massless, while the gauge fields along the broken generators become massive (again, in unitary gauge).
To ensure you understand this procedure, you should carry out this line of analysis for various other examples. I've shown you how the analysis goes for an $\mathrm{SU(2)}$ gauge theory Higgsed by an adjoint. You can come up with other examples of gauge groups and representations and work out the details, or study the many examples presented in the many field theory books out there.
By the way, the Georgi-Glashow model refers to an $\mathrm{SU(5)}$ gauge theory, not an $\mathrm{SU(2)}$ gauge theory.
Best Answer
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.