How do metals split the oxygen double bond during corrosion?
The corrosion of metals like iron is essentially an electrochemical process. 

In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of $\ce{Fe}$ (commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite complex, but it may be considered essentially as an electrochemical phenomenon. At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode and we can write the reaction.

\begin{align}
\tag{Anode}
  \ce{2Fe (s) &amp;-&gt; 2 Fe^{2+} + 4 e-} &amp;
  E^\circ_{\ce{Fe^{2+}/Fe}} &amp;= \pu{– 0.44 V}
\end{align} 

Electrons released at the anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of $\ce{H+}$ (which is believed to be available from $\ce{H2CO3}$ formed due to dissolution of carbon dioxide from air into water. Hydrogen ions in water may also be available due to dissolution of other acidic oxides from the atmosphere). This spot behaves as cathode with the reaction

\begin{align}
\tag{Cathode}
  \ce{O2(g) + 4 H+(aq) + 4 e- &amp;-&gt; 2 H2O (l)} &amp;
  E^\circ_{\ce{H^+/O2/H2O}} &amp;= \pu{1.23 V}
\end{align}

The overall reaction being:

\begin{align}
  \ce{2Fe(s) + O2(g) + 4H+(aq) &amp;-&gt; 2Fe^{+2}(aq) + 2 H2O (l)} &amp;
  E^\circ_{\text{cell}} &amp;= \pu{1.67 V}
\end{align}

The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide $\ce{(Fe2O3. x H2O)}$ and with further production of hydrogen ions.