Acidic cleavage of an ether
The Wikipedia page on ether cleavage says:


  Depending on the specific ether, cleavage can follow either $\mathrm{S_N1}$ or $\mathrm{S_N2}$ mechanisms. Distinguishing between both mechanisms requires consideration of inductive and mesomeric effects that could stabilize or destabilize a potential carbocation in the $\mathrm{S_N1}$ pathway. Usage of hydrohalic acids takes advantage of the fact that these agents are able to protonate the ether oxygen atom and also provide a halide anion as a suitable nucleophile.


So in essence, the mechanisms for such reactions are usually a trade-off between $\mathrm{S_N1}$ and $\mathrm{S_N2}$ pathways, with adequate carbocation stability playing a major role in the determination of the pathway (as intermediate cases like secondary cations often lead to a "confusing" case where it is difficult to say qualitatively whether the path is majorly $\mathrm{S_N1}$ or $\mathrm{S_N2}$).

In particular, the Wikipedia page cited above has pretty much answered your question under the $\mathrm{S_N2}$ ether cleavage tab:


  If the potential carbocation can not be stabilized, ether cleavage follows a bimolecular, concerted $\mathrm{S_N2}$ mechanism. In the example, the ether oxygen is reversibly protonated. The halide ion (here bromide) then nucleophilically attacks the less sterically hindered carbon atom, thereby forming methyl bromide and 1-propanol.
  


So for your case, the upper oxygen after protonation can cleave the $\ce{C-O}$ via a $\mathrm{S_N1}$ pathway to generate a reasonably stable potential carbocation which is delocalized over the benzene ring. 
Hence,it can simply remove a molecule of methanol(which is a good leaving group by itself) on it's own without needing an iodide to assist in the removal.

The strongly acidic solution of $\ce{HI}$ will anyway ensure it that the methanol undergoes $\mathrm{S_N2}$ to form methyl iodide later. Also, from the point of view of a nucleophile,it would be much easier to perform an $\mathrm{S_N2}$ reaction on a separated methanol molecule than on the entire substrate molecule initially given in the question due to steric considerations,so that also becomes a contributing factor.

But as for the lower oxygen,the poor fellow does not have such a choice. The potential carbocation that can be generated by cleaving the lower $\ce{C-O}$ bond is a phenyl cation,which is pretty unstable as explained here. So, like the Wikipedia page said and illustrated in the example above, the pathway has to be a concerted $\mathrm{S_N2}$ mechanism now so as to take the reaction forward. 

Note: Even if the said $\mathrm{S_N2}$ starts out slow, Le Chatlier's principle will ensure that ultimately, most of  the reactant goes into the product as equilibrium is attained.