Esters of catechol - Fries rearrangement
This Wikipedia article provides a nice overview of the Fries rearrangement. They note that the mechanism of the rearrangement is not fully understood, but the following mechanism is generally accepted.



The key step in the reaction is thought to involve the interaction of a lewis acid with a phenyl ester to generate an acylium ion ($\ce{[R-C=O]^+}$).  Once the acylium ion is generated the reaction proceeds like any other electrophilic aromatic substituion. The phenolic oxygen is a strong ortho-para director, so both ortho- and para-products are produced by the same mechanism. The $\ce{OCOR}$ group does not play a role in "directing" the substitution pattern. 

The ratio of ortho/para isomers in most aromatic electrophilic substitution reactions is very temperature dependent. This is because of kinetic vs. thermodynamic control. Typically, the para product is the kinetic product (first formed), while the ortho product is the thermodynamically more stable product and will form if the reaction is run at higher temperatures and allowed to equilibrate. For example, when phenol is mono-brominated at low temperature, the para-bromo product is formed in >95% yield (reference). So your specific reaction is likely being run under kinetically controlled conditions to produce the para product as the major product. As the Wikipedia link points out, the para product can equilibrate at higher temperatures to produce more of the ortho product.

Finally, just a word to address @bon 's comment.  Indeed, Fries rearrangement of catechol diesters does produce mono-acyl catechols (reference, p A-172) - apparently one of the acyl groups is lost in the process.