Proton removal during benzyne formation in substituted haloarenes
I appreciate Waylander's answer, but it did not address the OP's curiocity of why the base abstracts the proton from only ortho-position. Yet, Waylander correctly pointed out that since the aromatic $\pi$-system is at right angle to the triple bond (the second $\pi$-bond of the triple bond), the ability to donate a lone pair to the ring by a substituent (like $\ce{OCH3}$ here), or accept a lone pair from the aromatic nucleus (like $\ce{NO2}$ would do) is not relevant in this case. However, inductive effects of the substituents in aromatic nucleus are still relevant (i.e. electron-withdrawing effects that occur through single $\sigma$-bonds).
However, $\ce{t-BuO-}$ is not as strong a base as $\ce{H2N-}$ in usual cases of benzyne reactions. Thus, it would abstract relatively most acidic available proton from to aromatic nucleus. Compared to para-position, ortho-position is more acidic due to proximity (ortho is one $\ce{C-C}$ bond away compared to para which is three $\ce{C-C}$ bonds away). Following is an example you can find in literature (Ref.1):

References:

José-Antonio García-López and  Michael F. Greaney, &quot;Synthesis of biaryls using aryne intermediates,&quot; Chem. Soc. Rev. 2016, 45(24), 6766-6798 (DOI: https://doi.org/10.1039/C6CS00220J).