A conservation law is always due to global (not local!) gauge symmetries. Since a local symmetry implies a corresponding global symmetry, you also have a conservation law (in this case of a charge-like quantity).
Let's consider the most imple special case of the Noether theorem, which is just taylored to charge conservation. Here you have a symmetry, where the variation of the action
$$A[\psi]=\int \mathrm{d}^4 x \mathcal{L}(\psi,\partial_{\mu} \psi)$$
is unchanged under a socalled "external symmetry", i.e., a symmetry transformation not involving the space-time coordinates. For an infinitesimal transformation you have
$$\delta x^{\mu}=0, \quad \delta \psi=\delta \eta \tau(x,\psi),$$
where ##\tau(x,\psi)## is some function.
Now the variation of the action is
$$\delta A=\delta \eta \int \mathrm{d}^4 x \left [\frac{\partial \mathcal{L}}{\partial \psi} \tau + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \partial_{\mu} \tau \right] = \delta \eta \int \mathrm{d}^4 x \tau \left [\frac{\partial \mathcal{L}}{\partial \psi} \ - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \right]=0.$$
This means we have a symmetry if there exists a vector field ##j^{\mu}## such that
$$\left [\frac{\partial \mathcal{L}}{\partial \psi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \right] \tau=-\partial_{\mu} j^{\mu}. \qquad (*)$$
The sign on the right-hand side is convention.
The equations of motion are given by the Euler-Lagrange equations, i.e., the left-hand side vanishes for the solutions of the equations of motion, implying the continuity equation ##\partial_{\mu} j^{\mu}=0##, which means that the Noether charge
$$Q=\int \mathrm{d}^3 \vec{x} j^0(t,\vec{x})=\text{const}$$
and that this is a Lorentz scalar.
Now we need to find the explicit expression for ##j^{\mu}## given the Lagrangian and ##\tau##. To that end we rewrite (*) a bit:
$$\partial_{\mu} j^{\mu}=\partial_{\mu} \left [\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \tau \right]-\tau \frac{\partial \mathcal{L}}{\partial \psi} - \frac{\partial L}{\partial (\partial_{\mu} \psi)} \partial_{\mu} \tau.$$
So if we have a symmetry, we must have ##\Omega^{\mu}(x,\psi)## such that
$$\partial_{\mu} \Omega^{\mu}=-\delta L/\delta \eta=-\tau \frac{\partial \mathcal{L}}{\partial \psi} - \frac{\partial L}{\partial (\partial_{\mu} \psi)} \partial_{\mu} \tau. \qquad (**)$$
Here we have used that the symmetry transformation for the Lagrangian must not change the variation of the action for all fields, so it must be a total four-divergence of a function of only the fields and not its gradient. Then obviously
$$j^{\mu}=\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \tau+\Omega^{\mu}.$$
As the most simple example let's take the free Dirac field with the Lagrangian
$$\mathcal{L}=\overline{\psi}(\mathrm{i} \not{\partial}-m) \psi.$$
In this case the Lagrangian is obviously invariant under a phase transformation, i.e.,
$$\psi \rightarrow \psi'=\exp(-\mathrm{i} q \eta)\psi, \quad \overline{\psi} \rightarrow \overline{\psi}'=\exp(+\mathrm{i} q \eta) \psi,$$
where ##q## and ##\eta## are real numbers. Now we make ##\eta \rightarrow \delta \eta## small, getting
$$\delta \psi=-\mathrm{i} q \psi \delta \eta, \quad \delta \overline{\psi} = +\mathrm{i} q \overline{\psi} \delta \eta.$$
Thus we have
$$\tau=-\mathrm{i} q \psi, \quad \overline{\tau}=q \mathrm{i} \overline{\psi}.$$
Now
$$\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \psi )}=\overline{\psi} \mathrm{i} \gamma^{\mu}, \quad \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \overline{\psi})}=0.$$
To determine ##\Omega## we have to use (**), which in our case reads
$$\partial_{\mu} \Omega^{\mu}=0,$$
i.e., we can set ##\Omega^{\mu}=0## and thus the Noether Current is
$$j^{\mu}=q \overline{\psi} \gamma^{\mu} \psi.$$
It is easy to check that this is indeed a conserved current for the solutions of the equations of motion, i.e., the Dirac equation
$$(\mathrm{i} \not{\partial} +m) \psi=0.$$
To make the symmetry local, i.e., to make the Lagrangian invariant under a transformation of the form
$$\psi \rightarrow \psi'=\exp(-\mathrm{i} q \tau(x)) \psi, \quad \overline{\psi} \rightarrow \overline{\psi}'=\exp(+\mathrm{i} q \tau(x)) \overline{\psi},$$
we have to introduce a gauge field ##A^{\mu}## and plug everywhere where we have a partial derivative the gauge-covariant derivative
$$\partial_{\mu} \rightarrow \partial_{\mu} +\mathrm{i} q A_{\mu}.$$
If we define the transformation for ##A_{\mu}## as
$$A_{\mu}'=A_{\mu} + q \partial_{\mu} \tau$$
we get
$$\mathrm{D}_{\mu}' \psi'=(\partial_{\mu} + \mathrm{i} q A_{\mu}') \psi '=\exp(-\mathrm{i} q \tau)[\partial_{\mu} \psi-\mathrm{i} q \partial_{\mu} \tau \psi + \mathrm{i} q A_{\mu} \psi + \mathrm{i} q \partial_{\mu} \tau \psi]=\exp(-\mathrm{i} q \tau) [\partial_{\mu} \psi + \mathrm{i} q A_{\mu}]=\exp(-\mathrm{i} q \tau) \mathrm{D}_{\mu} \psi.$$
Thus now indeed the Lagrangian is invariant under local gauge transformations. Of course, the current is still conserved, because the Lagrangian is still invariant under global gauge transformations.
