Why are tertiary carbon atoms of polymers more prone to photo-oxidation and -degradation?
This is admittedly a very brief and somewhat tempting answer and it might not be complete. 

Photo-oxidation and -degradation occurs mainly via a radical mechanism. This is outlined on Wikipedia in more detail. In principle the initial step produces bond cleavage via UV radiation and with it radicals. Let $\ce{[P]_{$n$}}$ be a polymer, then $\ce{.[P]_{$n'$}}$ and $\ce{.[P]_{$n''$}}$ are decomposition products of this polymer, both of radical nature.
$$\ce{P -&gt;[h\nu] .[P]_{$n'$} + .[P]_{$n''$}}$$

In a saturated non chained polymer a homolytical bond cleavage would lead two primary radicals, i.e. $\ce{R-CH2.}$, which are very unstable. In a chained polymer, like PP, if the bond break occurs at the tertiary carbon, there would be a secondary radical $\ce{R-\overset{\bullet}{C}H-CH3}$ and another primary radical. Homolytical bond cleavage of a hydrogen carbon bond can even lead to a tertiary radical, $\ce{R-\overset{\bullet}{C}(CH3)2}$.  All in all this can lead to a higher concentration of radicals, and therefore more access points for oxygen to attack, hence propagating the chain. One of the driving forces is the gain in entropy of this process.

$$\begin{align}
\ce{.[P]_{$n''$} + O2 &amp;-&gt; [P]_{$n''$}OO.}\\
\ce{[P]_{$n$} + [P]_{$n''$}OO. &amp;-&gt; [P]_{$n''$}OOH + .[P]_{$n$}}\\
\ce{[P]_{$n''$}OOH &amp;-&gt;[h\nu] [P]_{$n''$}O. + .OH}\\
\ce{[P]_{$n$} + [P]_{$n''$}O. &amp;-&gt; [P]_{$n''$}OH + .[P]_{$n$}}\\
\ce{[P]_{$n$} + .OH &amp;-&gt; H2O + .[P]_{$n$}}\\
\text{etc.}&amp;\phantom{\ce{-&gt;}~~}\text{pp.}
\end{align}$$