Q:

How to compute this distribution? 

My question refers to this answer. 
I was hoping someone could explain in more detail the following reasoning. 

It remains to observe that $\Delta v$ is the distribution composed of 

the linear measure on $\partial Q$ 
$-\sqrt{2}$ times the linear measure on the diagonals of $Q$ 

This follows from considering the discontinuities of the normal
  derivative of $v$ across the aforementioned lines; elsewhere $v$ is
  harmonic.  One can also save the trouble of calculating the factor of
  $-\sqrt{2}$ by using the fact that $\int_{\mathbb R^2}\Delta v=0$.

I do not see how these measures arise from the discontinuities of the derivative.  I do see, however, why it is zero elsewhere. ( Because it is harmonic there, so the Laplacian vanishes.) 
I am familiar with the theory of distributions as outlined in Folland's book, so I don't need any formation about what distributions are.  I just want to know how to compute this specific one.  

A:

Consider the cube given $Q=[-1,1]^2$ and the diagonals $D_1,D_2$. These divide the cube into four triangles, call them $A_i$. Now we have for any $u\in C_c^\infty(\mathbb{R}^2)$ that
$$
\left\langle \Delta v, u \right\rangle = \int_{\mathbb{R}^2} v\Delta u dx = -\int_{Q} \nabla v\cdot \nabla u dx
$$
where the first equallity is by definition and the second is an integration by parts (made possible since $v$ is Lipschitz and is supported in $Q$).  Now we rewrite this last integral as
$$
\int_Q \nabla v\cdot\nabla u dx = \sum_{i=1}^4 \int_{A_i} \nabla v \cdot \nabla u dx \tag{1}
$$
We can apply again the divergence theorem (or Green's formula) to this since $v$ is smooth in each $A_i$, but now we have to take into account the boundary terms: In other words we get
$$
\int_{A_i} \nabla v\cdot \nabla u dx = -\int_{A_i} (\Delta v)u + \int_{\partial A_i} \frac{\partial v}{\partial \nu_i} u \sigma =\int_{\partial A_i} \frac{\partial v}{\partial \nu_i} u d \sigma ,\tag{2}
$$
where $\nu_i$ is the outward unit normal to $\partial A_i$ and $\sigma$ is the surface (or in this case length) measure.  
Order the $A_i$ so that $\nabla v=(1,0), (0,1), (-1,0), (0,-1)$ in $A_1,A_2,A_3,A_4$ respectively (in other words $A_1$ is the  triangle to the left and you enumerate counterclockwise).  Now notice that if $T$ represents a rotation of $\pi/2$ then 
$$
T( \nabla v|_{A_i}) = \nabla v |_{A_{i+1}} \quad \text{and} \quad T(\nu_i)=\nu_{i+1}
$$
With these we get that $\frac{\partial v}{\partial \nu_i} = \frac{\partial v}{\partial \nu_j}$ for every $i,j$. Take then $i=1$ and we compute that this derivative is $-1$ in $\partial A_1 \cap \partial Q$ and $1/\sqrt{2}$ in $\partial A_1 \cap Q$. With this we get
$$
\int_{\partial A_i} \frac{\partial v}{\partial \nu_i} u d \sigma = -\int_{\partial A_i \cap Q} u d \sigma + \frac{1}{\sqrt{2}}\int_{\partial A_i\cap Q} u d \sigma. 
$$
Adding these we get, noting that each of the two line segements in $\partial A_i\cap Q$ is shared with exactly one other $\partial A_j$ we get
$$
\sum_{i=1}^4 \int_{\partial A_i \cap Q} \frac{\partial v}{\partial \nu_i} u d \sigma = \sqrt{2}\sum_{i=1}^2\int_{D_i} u d \sigma. 
$$
In a similar fashion
$$
\sum_{i=1}^4\int_{\partial A_i \cap \partial Q} \frac{\partial v}{\partial \nu_i} u d \sigma = -\int_{\partial Q} u d \sigma. 
$$
This, together with (1) and (2), give
$$
\left\langle \Delta v, u \right\rangle = \int_Q u d \sigma - \sqrt{2} \int_{D_i} u d \sigma 
$$
which is what you want.