Here's what I perceive to be a mathematically and logically precise presentation of the theorem, let me know if this helps.
Mathematical Preliminaries
First let me introduce some precise notation so that we don't encounter any issues with "infinitesimals" etc. Given a field $\phi$, let $\hat\phi(\alpha, x)$ denote a smooth one-parameter family of fields for which $\hat \phi(0, x) = \phi(x)$. We call this family a deformation of $\phi$ (in a previous version I called this a "flow"). Then we can define the variation of $\phi$ under this deformation as the first order approximation to the change in $\phi$ as follows:
Definition 1. (Variation of field)
$$
\delta\phi(x) = \frac{\partial\hat\phi}{\partial\alpha}(0,x)
$$
This definition then implies the following expansion
$$
\hat\phi(\alpha, x) = \phi(x) + \alpha\delta\phi(x) + \mathcal O(\alpha^2)
$$
which makes contact with the notation in many physics books like Peskin and Schroeder.
Note: In my notation, $\delta\phi$ is NOT an "infinitesimal", it's the coefficient of the parameter $\alpha$ in the first order change in the field under the deformation. I prefer to write things this way because I find that it leads to a lot less confusion.
Next, we define the variation of the Lagrangian under the deformation as the coefficient of the change in $\mathcal L$ to first order in $\alpha$;
Definition 2. (Variation of Lagrangian density)
$$
\delta\mathcal L(\phi(x), \partial_\mu\phi(x)) = \frac{\partial}{\partial\alpha}\mathcal L(\hat\phi(\alpha, x), \partial_\mu\hat\phi(\alpha, x))\Big|_{\alpha=0}
$$
Given these definitions, I'll leave it to you to show
Lemma 1.
For any variation of the fields $\phi$, the variation of the Lagrangian density satisfies
\begin{align}
\delta\mathcal L
&= \left(\frac{\partial \mathcal L}{\partial\phi} - \partial_\mu\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\right)\delta\phi + \partial_\mu K^\mu,\qquad K^\mu = \frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\delta\phi
\end{align}
You'll need to use (1) The chain rule for partial differentiation, (2) the fact $\delta(\partial_\mu\phi) = \partial_\mu\delta\phi$ which can be proven from the above definition of $\delta\phi$ and (3) the product rule for partial differentiation.
Noether's theorem in steps
Let a particular flow $\hat\phi(\alpha, x)$ be given.
Assume that for this particular deformation, there exists some vector field $J^\mu\neq K^\mu$ such that
$$
\delta\mathcal L = \partial_\mu J^\mu
$$
Notice, that for any field $\phi$ that satisfies the equation of motion, Lemma 1 tells us that
$$
\delta \mathcal L = \partial_\mu K^\mu
$$
Define a vector field $j^\mu$ by
$$
j^\mu = K^\mu - J^\mu
$$
Notice that for any field $\phi$ satisfying the equations of motion steps 2+ 3 + 4 imply
$$
\partial_\mu j^\mu = 0
$$
Q.E.D.
Important Notes!!! If you follow the logic carefully, you'll see that $\delta \mathcal L = \partial_\mu K^\mu$ only along the equations of motion. Also, part of the hypothesis of the theorem was that we found a $J^\mu$ that is not equal to $K^\mu$ for which $\delta\mathcal L = \partial_\mu J^\mu$. This ensures that $j^\mu$ defined in the end is not identically zero! In order to find such a $J^\mu$, you should not be using the equations of motion. You should be applying the given deformation to the field and seeing what happens to it to first order in the "deformation parameter" $\alpha$.
Addendum. 2020-07-02 (Free scalar field example.)
A concrete example helps clarify the theorem and the remarks made afterward. Consider a single real scalar field $\phi:\mathbb R^{1,3}\to\mathbb R$. Let $m\in\mathbb R$ and $\xi\in\mathbb R^{1,3}$, and consider the following Lagrangian density and deformation (often called spacetime translation):
$$
\mathcal L(\phi, \partial_\mu\phi) = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi - \frac{1}{2}m^2\phi, \qquad \hat\phi(\alpha, x) = \phi(x + \alpha\xi)
$$
Computation using the definition of $\delta\mathcal L$ (plug the deformed field into $\mathcal L$, take the derivative with respect to $\alpha$, and set $\alpha = 0$ at the end) but without ever invoking the equation of motion (Klein-Gordon equation) for the field gives
$$
\delta \mathcal L = \partial_\mu(\xi^\nu\delta^\mu_\nu \mathcal L), \qquad \frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\delta\phi = \xi^\nu\partial_\nu\phi\partial^\mu\phi
$$
It follows that
$$
J^\mu = \xi^\nu\delta^\mu_\nu \mathcal L, \qquad K^\mu = \xi^\nu\partial_\nu\phi\partial^\mu\phi
$$
and therefore
$$
j^\mu = \xi^\nu(\partial_\nu\phi\partial^\mu\phi -\delta^\mu_\nu\mathcal L)
$$
If e.g. one chooses $\tau > 0$ and sets $\xi = (\tau, 0, 0, 0)$, then the deformation is time translation, and conservation of $j^\mu$ yields conservation of the Hamiltonian density associated with $\mathcal L$ as the reader can check.
Suppose, instead, that in the process of computing $\delta \mathcal L$, one were to further invoke the following equation of motion which is simply the Euler-Lagrange equation for the Lagrangian density $\mathcal L$:
$$
\partial^\mu\partial_\mu\phi = -m^2\phi,
$$
Then one finds that
$$
\delta\mathcal L = \partial_\mu(\xi^\nu\partial_\nu\phi\partial_\mu\phi)
$$
so $J^\mu = K^\mu$ and therefore $j^\mu = 0$, which is uninformative.
Firstly, you ask
isn't there more than one symmetry meaning this question has several answers?
Yes! In general, a given theory can have all sorts of symmetries, and each of these symmetries leads to its own conserved quantity via Noether's theorem.
As for what's going on with Noether's theorem and applying it in general, I'd like to strongly encourage you to read my answer here:
Noether's current expression in Peskin and Schroeder
If you still have questions after that, then let me know in the comments to this answer, and I'll gladly add an addendum to address your remaining confusions/questions.
Best Answer
For the given case since $\partial _\mu J^{\mu} = 0$, there is no need of adding this boundary term to the conserved current $j^{\mu}$ as the addition of it is meaningless as it won't play any part except adding up to a meaningless factor. We already have for the equation of the conserved current $\partial_{\mu} j^{\mu} = 0$ and we can choose to add any term $a^{\mu}$ to $j^{\mu}$ as long as it satisfies $\partial _{\mu} a^{\mu} = 0$. Hence it is really irrelevant to have the $J^{\mu}$ part.
As for the second part of the question, you are right in saying that the minus sign can be omitted as it is defined upto a multiplicative constant.