We can write the Fourier transform of
$\langle 0|\mathcal{T}A_{\nu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle$
as $$S(p) D_{\nu\alpha}(q) \ e\,\Gamma^{\alpha}(p,q,p+q)S(p+q)$$ where $S(p)$ is the full fermion propagator, $D_{\nu\alpha}(q)$ is the full photon propagator, $\Gamma^{\alpha}(p,q,p+q)$ is the proper vertex function, and an overall momentum conservation delta function has been dropped. Similarly, we can write the Fourier transform of $\langle 0|\mathcal{T}j^{\mu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle
$ as
$$S(p)V^{\mu}(p,q,p+q)S(p+q)$$ where $V^{\mu}(p,q,p+q)$ is a vertex function that we want to relate to $\Gamma^{\mu}(p,q,p+q).$ The vertex function $V^{\mu}(p,q,p+q)$ enters into the derivation of the Ward-Takahashi identity in Peskin and Schroeder on page 311, but the Ward-Takahashi identity is normally stated in terms of $\Gamma^{\mu}(p,q,p+q)$. Your conundrum (as I understand it) is that according to your analysis of the Schwinger-Dyson equation, $V^{\mu}(p,q,p+q)$ and $\Gamma^{\mu}(p,q,p+q)$ ought to differ by a factor of $Z_3$, but this contradicts the usual statement of the Ward-Takahashi identity where no such factor of $Z_3$ appears. I will argue from the Schwinger-Dyson equation that the longitudinal parts (in $q^{\mu}$) of $V^{\mu}(p,q,p+q)$ and $\Gamma^{\mu}(p,q,p+q)$ are equal, but that the transverse parts differ by the factor of $Z_3$ that you have found. Since only the longitudinal part enters into the Ward-Takahashi identity, the factor of $Z_3$ does not inter into that identity. You may want to review page 246 of Peskin and Schroeder. There they show that only the transverse part of the photon propagator is modified by the self-energy, but that in calculating Feynman diagrams we can simplify the analysis by including the self-energy in the longitudinal part as well because the longitudinal part does not contribute to the Feynman diagrams due to the Ward identity. However, the Schwinger-Dyson equation involves an inverse propagator which does not arise in Feynman diagrams and we need to reevaluate where the self-energy does and does not enter.
Specializing the Schwinger-Dyson equation to the case of $\langle 0|\mathcal{T}A_{\nu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle$ and Fourier transforming, we have
$$\tag{1} (D^{(0)\mu\nu}(q))^{-1} D_{\nu\alpha}(q) S(p) \ e\,\Gamma^{\alpha}(p,q,p+q)S(p+q) = \\ e\, S(p)V^{\mu}(p,q,p+q)S(p+q)$$
where $(D^{(0)\mu\nu}(q))^{-1}$ is the inverse of the non-interacting photon propagator. The Dyson equation for the photon propagator is
$$\tag{2} D_{\nu\alpha}(q) = D^{(0)}_{\nu\alpha}(q) + D^{(0)}_{\nu\beta}(q) i\Pi^{\beta\gamma}(q)D_{\gamma\alpha}(q) ,$$
so
$$\tag{3} (D^{(0)\mu\nu}(q))^{-1} D_{\nu\alpha}(q) = \delta^{\mu}_{\alpha} + i\Pi^{\mu\gamma}(q)D_{\gamma\alpha}(q).$$
Equation (1) then implies
$$\tag{4}\Bigl(\delta^{\mu}_{\alpha} + i \Pi^{\mu\gamma}(q)D_{\gamma\alpha}(q)\Bigr) \Gamma^{\alpha}(p,q,p+q) = V^{\mu}(p,q,p+q).$$
The Ward identity forces the longitudinal part of $\Pi^{\mu\gamma}(q)$ to vanish; that is, $q_{\mu}\Pi^{\mu\gamma}(q) = 0.$
Contracting equation (4) with $q_{\mu}$,
we therefore have
$$\tag{5} q_{\alpha} \Gamma^{\alpha}(p,q,p+q) = q_{\mu}V^{\mu}(p,q,p+q)$$
so no factor of $Z_3$ appears between the longitudinal parts of $\Gamma^{\alpha}(p,q,p+q)$ and $V^{\mu}(p,q,p+q)$ and therefore no factor of $Z_3$ appears in the Ward-Takahashi identity.
The transverse component does not enter the Ward identity for the vertex function but it is useful to consider the transverse component to illustrate where the factor of $Z_3$ does arise. Define $\Pi(q^2)$ by the equation
$\Pi^{\mu\nu}(q) = q^2(g^{\mu\nu} - q^{\mu}q^{\nu}/q^2)\Pi(q^2).$ The quantity $(g^{\mu\nu} - q^{\mu}q^{\nu}/q^2)$ can be described as a projection operator that projects out the transverse part of a vector.
Contracting equation (4) with $(g_{\nu\mu} - q_{\nu}q_{\mu}/q^2)$ and using the fact that $\Pi^{\mu\gamma}(q)$ is already transverse, we have
$$\tag{7} \Bigl(g_{\nu\alpha} - q_{\nu}q_{\alpha}/q^2 + i q^2\Pi(q^2)D^T_{\nu\alpha}(q)\Bigr) \Gamma^{\alpha}(p,q,p+q) =\\ \bigl(g_{\nu\mu} - q_{\nu}q_{\mu}/q^2\bigr)V^{\mu}(p,q,p+q),$$
where $$D^T_{\nu\alpha}(q) = \frac{-i}{q^2 (1-\Pi(q^2))} \bigl(g_{\nu\alpha} - q_{\nu}q_{\alpha}/q^2\bigr)$$ is the transverse part of the photon propagator (see page 246, Peskin and Schroeder). Equation (7) can then be written
$$\tag{8} \bigl(g_{\nu\alpha} - q_{\nu}q_{\alpha}/q^2\bigr) \Bigl(1/\bigl(1-\Pi(q^2)\bigr)\Bigr) \Gamma^{\alpha}(p,q,p+q) =\\ \bigl(g_{\nu\mu} - q_{\nu}q_{\mu}/q^2\bigr)V^{\mu}(p,q,p+q).$$
Now consider $q^2$ small enough that $\Pi(q^2)\approx \Pi(0)$ and use the relation (Peskin and Schroeder, page 246)
$$Z_3 = \Bigl(1/\bigl(1-\Pi(0)\bigr)\Bigr).$$ We have
$$\tag{9} \bigl(g_{\nu\alpha} - q_{\nu}q_{\alpha}/q^2\bigr) Z_3 \Gamma^{\alpha}(p,q,p+q) = \bigl(g_{\nu\mu} - q_{\nu}q_{\mu}/q^2\bigr)V^{\mu}(p,q,p+q).$$ So we see that the transverse parts of $V^{\mu}(p,q,p+q)$ and $\Gamma^{\mu}(p,q,p+q)$ differ by a factor of $Z_3.$
Before answering the full question, let me address the definition of $\mathcal{G}$ and connect to your last paragraph. For a generic principal $G$-bundle $\pi:P\to M$, the gauge group $\mathcal{G}(P)$ is defined as
$$\mathcal{G}(P) :=\left\{\vphantom{\int}\psi:P\to P\quad \middle|\quad \psi(p\, g) = \psi(p) g\quad \text{and}\quad \pi(\psi(p)) = \pi(p),\ ^\forall p\in P,\ ^\forall g\in G\right\}.\tag{1}$$
You can see that if $P$ is the trivial bundle $P=M\times G$, then the equation
$$\psi(p) = (m, \varphi(m) g), \qquad ^\forall (m,g)\in P,$$
defines for all $\psi\in\mathcal{G}(P)$ a function $\varphi:M\to G$ and thus, indeed
$$\mathcal{G}(M\times G) \cong \operatorname{Maps}(M,G)\equiv C^\infty(M,G)\equiv \Omega^0(M;G).$$
Now, going back to the generic case, the subgroup of $\mathcal{G}(P)$ generated by constant maps1 is precisely $G$, the group of global transformations:
$$ \left\{\text{constant}\ \psi\in\mathcal{G}(P)\right\}\cong G \subset \mathcal{G}(P).$$
These are the global transformations.
Now, to answer your questions, the Lagrangian and hence also candidates for Higgs-type potentials, should be $\mathcal{G}(P)$-invariant. You can show, for example, from the definition (1) that the Yang-Mills action is $\mathcal{G}(P)$-invariant. This should always be the case when you're building a gauge theory.
Moving on to your last bullet, let me first make an aside on terminology. Gauge invariance is not a symmetry, but a redundancy in the description. It can therefore not break. The scenario you describe there is most commonly referred to as having Higgsed the gauge invariance, and the term spontaneous symmetry breaking is reserved for global symmetries. Nevertheless, in the usual treatment of Higgsing, in order to arrive at the Higgsed Lagrangian it was necessary to fix a gauge i.e. choose a very carefully picked $\psi\in\mathcal{G}(P)$ and stick with it. You no longer have a gauge transformation available and the remaining symmetry is exactly $H\subset G$, not a larger, infinite group of bundle isomorphisms over it. This is explained (with a physics language) for example here.
1 in the trivial bundle case those are the ones that satisfy $\mathrm{d}\varphi = 0$, in the non-trivial case you can demand that patchwise and demand they glue nicely on overlaps
Best Answer
We do the exact same thing in the standard model when we break the electroweak symmetry. There are 3 broken generators $\delta_-/2, \sigma_1/2, \sigma_2/2$, and a leftover preserved $U(1)_{EM}$ generator, $\delta_+/2$, where
$$ \delta_\pm = \frac{1}{2}(\mathbf{1}\pm \sigma_3) $$
are linear combinations of the EW gauge group generators. You can verify that these generators span the same space as the original $SU(2)_L\times U(1)_Y$ generators.
You can simply just think of this as a choice of basis for the Lie Algebra that makes the leftover $U(1)$ symmetry manifest.