Building on the answer of Ján Lalinský and on further research, I add here my own answer to the question.

**question 1:** It turns out that the electric potential $V$ measured at some point of the wires identifies, up to a constant, to the electromagnetic potential $\varphi$ under the Coulomb gauge, which vanishes at infinity (this last condition specifies $\varphi$ uniquely in the whole space).  It can also be defined, up to a constant, as the integral of the conservative part of the electric field; this is the irrotational part of the E-field in the Helmholtz decomposition. 

To see why these two definitions are equivalent, let $\varphi$ be the potential under the Coulomb gauge $\nabla\cdot {\mathbf A}=0$. This last condition implies that $\mathbf A$ is a solenoidal field: ${\mathbf A} = \nabla \times \mathbf F$. 
We have Maxwell equation $${\mathbf E} = -\nabla \varphi - {\mathbf \partial_t A}
= -\nabla \varphi - \partial_t (\nabla \times {\mathbf F}) = -\nabla \varphi - \nabla \times (\partial_t \mathbf F),$$
hence $\nabla \varphi$ is the conservative part of $\mathbf E$.

Unfortunately, there is some of cheating here, because the "conservative part of the E-field" is not uniquely defined by Helmholtz decomposition which needs not be unique, even if it is specified as some "ground" point. Indeed, two possible functions $V_1$ and $V_2$ which agree at some point are possible: it suffices to choose an harmonic function $\Lambda$ equal to $0$ at the specified point, and to set $V_2 = V_1+\Lambda$. Then it is easily seen that if $\mathbf E = \nabla V_1 + \nabla\times \mathbf F_1$, then $\mathbf E = \nabla V_2 + \nabla \times (\mathbf F_1- \mathbf F_2)$, where $\nabla \times \mathbf F_2 = \nabla \Lambda$ (which is possible since $\nabla^2\Lambda = 0$ by definition).

So, it is probably better to define the electric potential $V$ as above, as the electromagnetic Coulomb potential $\varphi$ which vanishes at infinity, up to a constant. 


**question 2 and 3:**
Assume we have a  filiform two terminal device $D$ of some shape. Let $\rho(z, t)$ be the moving charge linear density at some point $z$ of $D$ and $\mathbf v$ its speed. Denote by $V_{emf}$ the integral of $d\vec\ell \cdot {\mathbf E + \mathbf v\times \mathbf B}$ along $D$, from the extremity where the current exits to the extremity where the current enters into the wire.

Pay attention that $V_{emf}$ is not defined as a potential *between two points*, but as an electromotive force *along a (filiform) device*. 
Since $\mathbf v\times \mathbf B$ is orthogonal to $\mathbf v$, and hence to $d\vec\ell$, We have  $$V_{emf} = \int_D E \cdot d\vec\ell .$$
Furthermore, the power transmitted by the electromagnetic field to $D$ is 
$$P = \int_D d\mathbf F\cdot \mathbf v = \int_D (\rho\, d\ell )\,({\mathbf E + \mathbf v\times \mathbf B})\cdot \mathbf v = \int_D \rho\, \mathbf E\cdot \mathbf v \,d\ell
= \int_D\mathbf E \cdot \mathbf j d\ell.$$ 
If we assume that the current is uniform inside $D$ which is a valid assumption as far as the wave length of the electric wave is large with respect to the dimension of the device, then $||\mathbf j|| = I$, $\mathbf j = I{d\vec \ell \over d\ell}$ and $$P = I\int_D \mathbf E\cdot d\vec\ell = I V_{emf}.$$
This is valid for every filiform two terminal devices.

On the other hand, one of the most basic law of electrical engineering is that the power transmitted to a two terminal device is $P = VI$, where $V$ is the electric potential between the two terminals. 
We see that this is valid only whenever $V = V_{emf}$, that is, with Maxwell equation $E = -\nabla \varphi - \partial_t \mathbf A$, whenever $\partial_t\mathbf A = 0$ (under the Coulomb gauge). Since $\mathbf B = \nabla \times \mathbf A$, this implies $$\partial_t \mathbf B = 0.$$
The meaning of this equation is :

*Electric laws of the form $\Delta V = f(I)$, related to two terminal devices, are valid only whenever there is no varying magnetic field along the device.*

Conversely, if this condition holds, then $\nabla \times \partial_t \mathbf A = 0$, hence $\partial_t A$ is irrotational: $\partial_t\mathbf A = \nabla \mathbf F$. Coulomb gauge $\nabla\cdot \mathbf A = 0$ implies $$\nabla^2 \mathbf F = 0.$$  
Observe that $\mathbf E$ vanishes at infinity (since the electric wires are contained in a bounded domain). From the definition of $\varphi$ and Maxwell equation
$$\mathbf E = -\nabla \varphi - \partial_t \mathbf A,$$
it follows that $\partial_t \mathbf A = \nabla \mathbf F$ vanishes at infinity. Hence $\mathbf F$ is constant (uniqueness of the solution of Laplace equations with Neuman boundary conditions). Hence $\partial_t \mathbf A = 0$.
We see that the above condition is also sufficient.

In general, laws of the form $V = f(I)$ for two terminal devices have to be replaced by 
$$V_{emf} = f(I),$$ and one has to be careful that $V_{emf}$ is not a potential between two points, that would be independant of the path joining those points, but an "electromotrive force" *along a given path*, which depends also on the shape of the path. 
If the path is oriented, then $V_{emf}$ is defined as above if the orientation is opposite to the current flow, and is defined as its opposite otherwise. 
With these conventions, everything holds true by replacing everywhere $V$ by $V_{emf}$, except for the path dependence. Nevertheless, at those zones of the circuit where there is no magnetic induction, the electromotive force $V_{emf}$ along the path joining two points $A$ and $B$ is equal to $V(B) - V(A)$, as explained above. So everything works, if one is careful to isolate those zones where there is some magnetic induction and to give them a special treatment.   

This works in synergy with induction loops (inductor etc.), where all what is specified is the electromotive force as a function of the flux $\Phi$:
$$V_{emf} = \partial_t \Phi.$$  

So, the apparent paradox in question 3 can be solved as follows:
$$V_{emf} = \partial_t \Phi = V_{emf}({\rm path_1}, R_1) + V_{emf}({\rm path_2}, R_2) = R_1 I + R_2 I = (R_1 + R_2)I.$$
Hence $I = V_{emf}/(R_1+R_2)$. 

At the terminals of the loop, and for any inductor in general, the magnetic induction is weak and we have $$V_{emf}(\mathrm{along\ any\ path\ joining\ the\ terminals}) = \Delta V$$
(assuming the other paths joining the terminals lie in induction free zones).