Effect of effective nuclear charge increase on the stabilization of 2s and 2p orbitals
2s orbitals are stabilized more than 2p orbitals by the effective nuclear charge because of better penetration.

Everything you wrote above is correct except for,


  2p orbitals are more stabilized because they penetrate less in the
  space of the 1s orbital


It is correct that they penetrate less, but they are not stabilized more.  Because they penetrate less, they are stabilized less by the effective nuclear charge.  Electron penetration correlates with electron stabilization.

Edit: Response to OP's first comment


  explain then why 2nd period non-metals can't easily form higher
  oxidation states while 3p elements can?


First off, second period non-metals can form higher oxidation states, for example $\ce{NO3^{-}}$ where the nitrogen is in the +5 oxidation state.  But I understand what you're getting at, and the answer comes back to the difference in effective nuclear charge that 2p and 3p electrons feel.  3p electrons are further from the nucleus and screened by all of the 2s and 2p electrons.  3p electrons are therefore held more loosely than 2p electrons.  This leads to lower electronegativities, 



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lower ionization potentials



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and higher oxidation states being more common for third period non-metals compared to second period non-metals. 

Edit: Response to OP's second comment


  3p electrons penetrate well into the space near the nucleus (link)
  Therefore it
  should be easier for 2p elements to form higher oxidation states than
  3p elements, but it's not, how?


Your link (Rose-Hulman) to the 2p-3p radial electron distribution is good, it shows that the 3p electrons are "on average" further from the nucleus than 2p electrons - which is the key point.  Here is a 3-dimensional representation of the same thing.  Look how much further away the 3p electrons are from the nucleus than the 2p electrons.  Given that the 3p electrons are 1) better screened (more interior electrons) and 2) much further away from the nucleus than the 2p electrons, it will much easier to remove more 3p electrons than 2p electrons and achieve higher oxidation states.  This is consistent with the electronegativity and ionization potential arguments presented in my first edit.



Edit: Response to OP's third comment


  the effective nuclear charge has bigger increase for 2p than 2s
  (except N-O) according to this chart. How so?


The key here is that the s orbital is spatially symmetric, it has a spherical pattern.  An electron in an s orbital can screen outer electrons from the nucleus equally in all directions.  On the other hand, a p orbital is directional, it is not spherically symmetric.  An electron in the 2px orbital will not screen outer electrons from the nucleus equally in all directions.

As we proceed from boron to carbon, that first 2s electron is effective in screening the second 2s electron because of its spherical orbit (2s effective nuclear charge increase=0.641).  However, the first 2p electron is not as effective in screening the nucleus because of its non-spherical shape.  So when we add the second 2p electron, it "feels" the nucleus more strongly than the second 2s electron did so the change in effective nuclear charge is greater for the second p electron than it was for the second 2s electron (2p effective nuclear charge increase=0.715).

You noticed the "anomaly" for the 2p electrons as we go from nitrogen to oxygen.  This is because with 3 2p electrons, all of the nitrogen 2p orbitals (px, py, pz) have an electron and produce a spherical electron distribution pattern.  The nucleus is better-screened by the 3 2p electrons in nitrogen, so when we add one more p electron to yield oxygen, the effective nuclear charge increase is noticeably smaller (N -> O 2p effective nuclear charge increase=0.619).