Where does the negative sign of the Laplacian in the four-divergence go? As far as I'm aware from my course notes and what I've found online, the four vector gradient expands as
$$
\begin{align*}
\partial_\mu x_\mu &= \left( \frac{1}{c} \frac{\partial}{\partial t}, -\frac{\partial}{\partial x^1}, -\frac{\partial}{\partial x^2}, -\frac{\partial}{\partial x^3} \right)\left( x_0, x_1, x_2, x_3 \right)\\
&= \left(\partial_0 x_0, -\vec{\nabla}\vec{x} \right)
\end{align*}
$$
where $c=1$,
and the four vector divergence expands as
$$\begin{align*}\partial_\mu x^\mu &= \left( \frac{1}{c} \frac{\partial}{\partial t}, -\frac{\partial}{\partial x^1}, -\frac{\partial}{\partial x^2}, -\frac{\partial}{\partial x^3} \right) \cdot \left( x^0, x^1, x^2, x^3 \right)\\
&= \left( \frac{1}{c} \frac{\partial}{\partial t}, - \vec{\nabla} \right) \cdot \left( x^0, \vec{x} \right)\\
&= \partial_0 x^0 + \vec{\nabla} \cdot \vec{x}
\end{align*}$$
In the latter case, where does the negative sign caused by the metric signature go?
 A: The divergence of the four-vector defining position  $x^\mu = (x^0, x^1,x^2,x^3)$  will give you the dimension of the spacetime and is given by the scalar product or $\partial_\mu$ with $x^\mu$. It is defined as 
$$ \partial_\mu x^\mu = \frac{\partial}{\partial x^0} x^0 + \left( \frac{\partial}{\partial x^i} x^i \right)  = 4,  $$
where we sum over repeated indices.
Or, for an arbitrary four-vector $a^\mu = (a^t, \mathbf{a})$ where $\mathbf{a}$ is the Cartesian spatial components is given by 
$$ \partial_\mu a^\mu = \frac{\partial}{\partial t} a^t  + \mathbf{\nabla} \cdot\mathbf{a}.   $$
Your confusion regarding the minus sign is that you will get one minus sign from the four-vector scalar product and the other comes from the spatial components of $\partial_\mu$. Hope that clears it up.
A: Your first expression is not well-defined. According to the Einstein summing convention, you sum over double indices, where one index is upper and the other one is lower. Such indices are called silent. Such an expression is then explicitly written as:
$$x^\mu y_\mu = x^0y_0 + x^1y_1 + x^2y_2 + x^3y_3 = x^0y^0 - x^1y^1 - x^2y^2 - x^3y^3$$
Note the sign change when passing from a lower to an upper index (works only in Minkowski metric).
Expressions like $x^\mu y^\mu$ or $x_\mu y_\mu$ are not well-defined, don't use them!
Indices which only appear once (on each side of an equation if an equation is considered) are called free indices. Example:
$$x^\mu y^\nu = ?$$
Since they are free, you need to specify some index values to evaluate the expression explicitly, say:
$$\left(x^\mu y^\nu\right)_{\mu=2\nu=3} = x^2y^3$$
You need to be careful when derivatives are involved. Derivatives are "naturally" defined with lower indices:
$$x^\mu = (x^0, \vec x),\qquad x_\mu = (x^0,-\vec x) $$
but
$$\partial_\mu = (\partial_0, \vec \nabla),\qquad \partial^\mu = (\partial^0,-\vec \nabla) $$
where $\vec x = (x^1, x^2, x^3)$, $\vec \nabla = (\partial_1, \partial_2,\partial_3)$ and the metric convention is that $x_0=x^0$ and $x_i=-x^i$.
Therefore,
$$x^\mu y_\mu = x^0 y^0 - \vec x\vec y$$
as above, and also
$$\square = \partial^\mu \partial_\mu = (\partial_0)^2 - {\vec \nabla}^2 $$
but
$$\partial_\mu x^\mu = \partial_0 x^0 + \vec \nabla \vec x$$
