Off-shell and on-shell assumptions within the derivation of Noether's theorem If we consider a transformation of a field $\Phi \rightarrow \Phi + \alpha \frac{\partial \Phi}{\partial \alpha}$ which is not a symmetry of a lagrangian then one can show that the Noether current is not conserved but that instead $\partial_{\mu}J^{\mu} = \frac{\partial L}{\partial \alpha}$.
I think the way this is derived is as follows $$\delta S = \int d^4 x \, \delta L = \int d^4 x \left( \left( \frac{\partial L}{\partial \Phi} \delta \Phi - \partial_{\mu} \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \right) \delta \Phi + \partial_{\mu} \left(\frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) \right)$$ Then the first term is zero due to the equations of motion and so we are left with the second total divergence term with $$J^{\mu} = \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \frac{\partial \Phi}{\partial \alpha}$$ so we are left with $$\delta S = \int d^4 x \, \alpha \partial_{\mu} J^{\mu}.$$ 
Writing out $$\delta L = \frac{\partial L}{\partial \Phi}\delta \Phi + \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta (\partial_{\mu} \Phi),$$ inserting $\delta \Phi = \alpha \, \partial \Phi/\partial \alpha$  we see that $$\delta L = \alpha \frac{\partial L}{\partial \alpha}.$$ Then we can compare with the above and deduce the result.
My questions are:
What permits the use of the equations of motion here? If the equations of motion hold then $\delta S = 0$ identically in that the solutions to such equations minimise the action. Using the equations of motion gives me $\int \partial_{\mu} J^{\mu} d^4 x = \delta S$ in the end as shown above but since I used the equations of motion isn't this just equal to zero? And also since we are always left with an integral of a total divergence isn't this always zero on the physical assumption that the field variations vanish at infinity/boundary of experiment?
I've seen the nice questions and answers posted e.g here 
https://physics.stackexchange.com/question/327999/  and the answer here by Qmechanic Which transformations *aren't* symmetries of a Lagrangian? 
Basically I'd like to understand what was said in that answer and see it in practice with the above non symmetry of the lagrangian.
 A: Let me present for you the derivation of Noether's theorem that I really like and that makes all assumptions of Noether's theorem clear. Let $\varphi$ be the set of fields in the theory so that
$$
S[\varphi] = \int d^4 x {\cal L}( \varphi(x), \cdots )
$$
Now, suppose we have a connected, continuous symmetry of the action $\varphi(x) \to \varphi_\alpha'(x)$ parameterized by a real number $\alpha \in {\mathbb R}$ so that $\alpha = 0$ corresponds to the identity transformation $\varphi'_{\alpha=0}(x) = \varphi(x)$. The claim that the transformation above is a symmetry of the action (see my answer to What constitutes a symmetry for Noether's Theorem?) is
$$
S[\varphi'_\alpha] = S[\varphi]
$$
Note that this symmetry is off-shell. 
Now, take $\alpha$ to be small and write
$$
\varphi'_{\alpha}(x) = \varphi(x) + \alpha \psi(x) + {\cal O}(\alpha^2) ~. 
$$
Then,
$$
S[\varphi + \alpha \psi] = S[\varphi]  + {\cal O}(\alpha^2) \qquad \qquad (1) 
$$
OK! So far, so good. Now, consider a different field redefinition $\varphi(x) \to \varphi'(x) = \varphi(x) + \alpha f(x)\psi(x)$. Note that, now since $f(x)$ is a function as opposed to a constant, the above field redefinition is not a symmetry of the action. This means that if I consider $S[\varphi + \alpha f \psi]$ there is in general an ${\cal O}(\alpha)$ unlike what's happening in (1). However, we know that this ${\cal O}(\alpha)$ term vanishes whenever $f(x)=$ constant since in that case, we would have a symmetry. This means that the ${\cal O}(\alpha)$ term must take the form
$$
S[\varphi + \alpha f \psi] = S[\varphi]  + \alpha \int d^4 x \partial_\mu f(x) j^\mu(x) + {\cal O}(\alpha^2) \qquad \qquad (2)  
$$
A derivative must act on $f(x)$ in the ${\cal O}(\alpha)$ term so that it vanishes when $f(x)=$ constant. 
Now, here comes the kicker. In (2), let $\varphi=\varphi_0$ where $\varphi_0$ be a solution to the equations of motion. The, by the variational principle
$$
S[\varphi_0+\alpha \delta \varphi] = S[\varphi_0] + {\cal O}(\alpha^2)
$$
Note that this is true for any variation $\delta \varphi$ (there are caveats to this as well, but we will not discuss this). We then find
$$
S[\varphi_0 + \alpha f \psi] = S[\varphi_0]  + \alpha \int d^4 x \partial_\mu f(x) j^\mu(x) + {\cal O}(\alpha^2) = S[\varphi_0] + {\cal O}(\alpha^2)
$$
which implies that if and only if $\varphi$ is taken to be on-shell, then
$$
\int d^4 x \partial_\mu f(x) j^\mu(x) = 0~. 
$$
This is true for any function $f(x)$ (subject to the same caveats mentioned previously). So we can take $f(x) = \delta^4(x-y)$ and the equation above then reduces to
$$
\partial_\mu j^\mu(y) = 0~. 
$$
Let us summarize all our assumptions. We assume that there exists a connected continuous global off-shell symmetry of the action that is generated by a real parameter. We have then proved that given this, there exists a current $j_\mu(x)$ which is conserved $\partial_\mu j^\mu(x)=0$ on-shell. 
A: It seems the heart of OP's question is the following. 

It is well-known that boundary conditions (BCs) are necessary in order to derived Euler-Lagrange (EL) equations. Does the infinitesimal transformations in Noether's theorem satisfy the pertinent BCs? 

Answer: This is typically not the case. So boundary terms (BTs) will not necessarily vanish. See also this related Phys.SE post. 
Example 1: Consider a free particle 
$$L~=~\frac{1}{2}m\dot{q}^2.$$ 
Infinitesimal off-shell translational symmetry 
$$\delta q~=~\epsilon$$ 
(which leads to on-shell momentum conservation via Noether's theorem) does not satisfy Dirichlet BCs
$$ q(t_i)~=~q_i\quad\text{and}\quad q(t_f)~=~q_f. $$
Example 2: Consider a harmonic oscillator
$$L~=~\frac{1}{2}m\dot{q}^2-\frac{1}{2}kq^2.$$
Infinitesimal translation transformation
$$\delta q~=~\epsilon$$ 
(which is not an off-shell quasi-symmetry) leads to the following variation of the Lagrangian
$$ \delta L~=~\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial\dot{q}}\frac{d}{dt}\delta q
~=~-(m\ddot{q}+kq)\delta q +\frac{d}{dt}(p\delta q)~\stackrel{\text{on-shell}}{\approx}~ \epsilon \frac{dp}{dt},  $$ 
which is not necessarily zero but guaranteed to be total time derivative on-shell. This corresponds to the well-known fact that momentum 
$$ p~:=~\frac{\partial L}{\partial\dot{q}}~=~m\dot{q} $$
is not conserved on-shell in the harmonic oscillator.
A: What's confusing you is that you've jumped the gun in your first equation, by racing ahead with the derivation of the Euler-Lagrange equations.
Start instead with
$$\delta S = \int d^4x \frac{\partial L}{\partial \phi}\delta\phi + \frac{\partial L}{\partial(\partial_\mu \phi)}\delta(\partial_\mu \phi)   $$
In the derivation of the Euler-Lagrange equations, we impose the principle of least action that $\delta S =0 $, then and integrate by parts to obtain your first equation. The boundary term can be assumed to vanish, but as $\delta \phi$ is arbitrary, we demand that the coefficient of $\delta \phi$ must vanish to satisfy the least action principle, which implies our equation of motion.
For a symmetry transformation $\delta \phi$ is not arbitrary; it is a symmetry of the action. This means that it leaves the action unchanged. As an example, a Lagrangian for a complex field 
$$L =\frac{1}{2}\partial_\mu \phi \partial^\mu\phi^* + \frac{1}{2}m^2 \phi^* \phi$$ has a symmetry $\phi \rightarrow e^{i\alpha} \phi$, corresponding to an infinitesmal variation $\delta\phi = i\alpha \phi$. This transformation leaves the Lagrangian $L$ unchanged; we don't need to use the equations of motion to enforce this. 
Speaking with generality, whilst the above transformation left $L$ itself invariant, $\delta \phi$ will be a symmetry of the action if it only modifies the Lagrangian by a divergence term $L \rightarrow L + \partial _ \mu j^{\mu} (x)$ which is assumed to vanish at infinity, so as to leave the action invariant. So 
$$\delta S = \int d^4x \frac{\partial L}{\partial \phi}\delta\phi + \frac{\partial L}{\partial(\partial_\mu \phi)}\delta(\partial_\mu \phi) =  \int d^4x  \partial _ \mu j^{\mu} (x) =0$$
- to repeat, not by the principle of least action, but by the definition of a symmetry. 
We can write, at this stage trivially,
$$ \int d^4x\left( \frac{\partial L}{\partial \phi}\delta\phi + \frac{\partial L}{\partial(\partial_\mu \phi)}\delta(\partial_\mu \phi) -  \partial _ \mu j^{\mu} (x)\right) =0$$
In fact, by our definition of $\partial _ \mu j^\mu$ as the local change in the Lagrangian $L$, we don't even need that integral sign there, but let's agree to integrate the functional derivatives by parts before we get rid of it. We thus find 
$$\left( \frac{\partial L}{\partial \phi}  - \partial_\mu \frac{\partial L}{\partial(\partial_\mu \phi)}\right)\delta \phi +  \partial_\mu\left(\frac{\partial L}{\partial(\partial_\mu \phi)}\delta \phi -j^{\mu} (x)\right) =0$$
 Now, let's go "on-shell" and think about what happens when the equations of motion are satisfied. Then the first term vanishes, so we infer that the current 
$$J^{\mu} =\frac{\partial L}{\partial(\partial_\mu \phi)}\delta \phi -j^{\mu} (x)$$ is conserved. 
Typically speaking, this current is not trivial, and you can check that it is conserved when $\phi$ satisfies its equation of motion. For the example above, you should find something like 
$$J^{\mu} = \phi^* \partial^\mu \phi -\phi \partial^\mu \phi^*$$
which vanishes if and only if $\phi$ satisfies the Klein-Gordon equation. (In the quantum theory, $\phi$ is a charged field, and $J^{\mu}$ is the current density.) 
To be clear about the relevance of the equations of motion in the conservation law, consider @Qmechanic 's example of a free particle above. Translation symmetry is a symmetry of the action; if you have some mad path from A to B that zig-zags every which way, its action will be equally ridiculous if you shift all the points on that path to the left by 5cm. But along that hypothetical zig-zag path momentum is not conserved; it does not obey the equation of motion. Instead, along the path that does satisfy the equation of motion (i.e. a straight line), momentum is conserved. I hope that helps clear things up.
