The enolate of p-chloropropiophenone --namely, B-- is being generated from ketone A by a kinetic base such as lithium diisopropylamide (LDA) in a polar, aprotic solvent, typically tetrahydrofuran (THF). You've showed the correct mechanism for obtaining the O-oxalylation product E. If E is formed at all, or if it were exposed to LiOEt independently, it would revert to enolate B and diethyl oxalate (DEO) via microscopic reversibility. The intermediate in this process is the one you have presented. Oxalylation of B can also occur on carbon, which leads to C. This acidic β-diketone is irreversibly deprotonated to form the enolate D and ethanol. Treatment of D with aqueous HCl would yield β-diketone C.
There are other subtleties in this reaction mechanism. If enolate B deprotonates β-diketone C forming ketone A and enolate D, can LiOEt deprotonate ketone A under the reaction conditions? If not, because the reaction requires stoichiometric base, one would require a 2/1 ratio of B/DEO or LDA/A.
But this part is probably TMI. Let's move on to pKa's.
The larger the pKa, the weaker the acid and the stronger the conjugate base. Approximate pKa values (logarithmic scale) of the "acids" in this reaction are listed above in decreasing order of conjugate base strength. This means that LDA deprotonates ketone A; LiOEt deprotonates β-diketone C; and enolate D deprotonates HCl. This process is known as the leveling effect.
A chemist would not do this reaction this way. Too much work! The low temperature preparation of LDA using n-butyllithium with syringes in an inert atmosphere is a bit of a hassle. It is much simpler to use sodium ethoxide in ethanol as the base and solvent. No inert atmosphere required and it is amenable to scale-up. Take a look at this Organic Synthesis procedure (https://DOI:2010.15227/orgsyn.011.0042) for the oxalylation of cyclohexanone. Take note of the stoichiometry of the reaction.
