Orbital correlation diagram for the cleavage of cyclopropylmethylene carbene
What is the type of the ring opening; is it a concerted or radical
  based C-C bond cleavage?


This is a chelotropic reaction, a subset of cycloaddition\fragmentation reactions.  As to whether it proceeds by a concerted or radical mechanism, well, you haven't presented any experimental data that can help us decide.  For example, if the reaction had been run with the cis-2,3-dimethylcyclopropane analogue and both cis- and trans-but-2-ene were found among the products, we could argue for a biradical, two-step (non-concerted) process.  When we create an orbital symmetry diagram, that just allows us to make a prediction as to whether or not we expect the reaction to be concerted.


  I strongly assume the empty p-orbital to be in hyperconjugation with
  the walsh orbitals


There are a number of different conformations available to the molecule, but I agree with you that having the carbenic p-orbital aligned with the plane of the cyclopropane ring (the "bisected" conformation) should be favored.  Just as the bent carbon-carbon bonds with high p-character in a cyclopropane ring stabilize the empty p-orbital in a cyclopropylcarbinyl cation (your drawing above), the same thing should happen here with the carbene's empty p-orbital.


  I was asked to draw the orbital correlation diagram.


I have difficulty reading your diagram, so I created a new orbital correlation diagram.



Let me say a few things about my figure in order to make it easier to understand.


The rear cyclopropane bond is bisected by the plane of the screen, this plane also contains the cyclopropyl carbon connected to the carbene center.  
The carbene p-orbital is perpendicular to the plane of the screen, the carbenic $\ce{sp^2}$ orbital lies in the plane.
The acetylenic sigma bond lies in the plane of the screen just as the sigma bond connecting the cyclopropyl carbon to the carbenic carbon does.
There are two pi bonds in the acetylene, Px is perpendicular to the plane and Pz is contained in the plane of the screen.
This plane of the screen is the plane of symmetry that we will use to determine the symmetry of the various orbitals and bonds involved in this transformation.
I haven't redrawn the Walsh cyclopropane orbitals, we can refer to your drawing up above.
$\mathrm{\sigma_1}$, $\mathrm{\sigma_2}$ and $\mathrm{\sigma_3}$ refer to the 3 bonding Walsh orbitals (cyclopropane) that you've drawn above.
I haven't drawn any of the antibonding orbitals, as we'll see below, we don't need them in this case.
The empty carbenic p-orbital and $\ce{sp^2}$ orbital should be around E=0, but I've drawn them slightly below for clarity.
The "A" and "S" labels tell us whether the orbital is antisymmetric or symmetric with respect to our preserved symmetry element, the plane of the screen.


Now, if you look at the reactant occupied orbitals and their symmetries, we have 3 orbitals with S symmetry and 1 with A symmetry.  In the products we also have 3 occupied orbitals with S symmetry and 1 with A symmetry. So there is a smooth correlation between reactant orbitals and product orbitals.  This analysis suggests that the thermal reaction can proceed in a concerted fashion.

Miscellaneous


Just because a reaction is allowed to proceed in a concerted fashion does not mean that the reaction will occur.  The most common reaction of cyclopropyl carbenes is ring expansion to the corresponding cyclobutene.  Other concerted reactions may be preferred because they happen to be of lower energy.
Of course, if a reaction is permitted in the forward direction, then it is also permitted in the back direction.  But when we heat ethylene and acetylene we don't generate cyclopropyl carbene - why not?  Well, these are all equilibria and the forward reaction is extremely exothermic.  We remove a highly strained cyclopropane and a high energy carbene and replace them with pi bonds.  Therefore we would expect the back reaction (combination of ehtylene and acetylene) to be extremely endothermic.  The products are much more stable than the reactants and we don't expect to see a significant concentration of reactants at equilibria.  It would be interesting to heat ethylene and acetylene to high temperature and see if any cyclobutene (or its ring opened product, buta-1,3-diene) is formed.  If these products are observed they would be suggestive of the presence of cyclopropyl carbene which then underwent the ring expansion reaction to cyclobutene.


Response to OP's comments:


  How can we see in general from an orbital symmetry correlation diagram
  if a reaction is concerted or not?


If all of the reactant bonding MO's correlate with product bonding MO's, then the reaction is allowed to proceed in a concerted fashion.  If some reactant bonding MO's correlate with antibonding product MO's, then the reaction is not allowed to proceed in a concerted fashion.

In the current example, the reactant bonding (occupied) MO's $\sigma_1,~ \sigma_2,~ \sigma_3$ and the carbene $\ce{sp^2}$ correlate with product MO's $\sigma$, ethylene $\pi$, acetylene $\pi_x$ and acetylene $\pi_z$ respectively.  All ground state reactant orbitals correlate with ground state product orbitals.  Therefore the reaction is allowed to proceed in a concerted manner.


  Is it okay to say biradicaloid instead of carbene?


No, not really. "Biradical" is generally used in cases where the electrons are on different atoms.  Like when we break a bond in cyclopropane we have a 1,3-biradical.


  In the orbital correlation diagram I would predict the acetylene a
  little bit lower because of more s character (sp) than ethylene (sp2).
  May be the energy difference is roughly as high as the difference
  between carbene p orbital A and carbene sp2 orbital.


That may be, but in both ethylene and acetylene the pi electrons (which is what we are talking about) are in pure p orbitals.  My guess was that they would have roughly the same energy.


  For the retro cheletropic reaction I assume a photolytical pathway


The retro reaction, like the forward reaction, should be thermally allowed and photochemically forbidden.


  According to the paper, I've found here...


I can only see the abstract, but from what I read, I don't think this is a reaction involving a photochemically produced carbene in an excited state.  I think that they are photochemically decomposing the diazo compound and produce the same carbene they generate thermally at higher temperatures.  The photochemical route just allows them to generate the same ground state carbene at very low temperatures.