General rule: do not use DEG_get_original_object in the evaluation functions. There are some exceptions when this is needed by design, but those are rare and isolated cases.

You should be able to use object data ID's recalc from the evaluated domain.

Can we extract this block to a separate function to keep BKE_object_eval_uber_data "clean"?

Also I think all this bit manipulation logic could be made more readable. Took me a while to understand what you are actually checking for here exactly. Maybe something like:

const IDRecalcFlag combined_recalc = ob->id.recalc | ((ID *)ob->data)->recalc;
const bool is_recalc_geometry = combined_recalc | ID_RECALC_GEOMETRY;
const bool is_recalc_deform = combined_recalc | ID_RECALC_GEOMETRY_DEFORM;

if (is_recalc_deform && !is_recalc_geometry) {
   BKE_object_data_batch_cache_deform_tag(ob->data);
}
else {
   BKE_object_data_batch_cache_dirty_tag(ob->data);
}

Note, the semantic here is not exactly the same as in your patch (As can be seen be the fact that you might not always tag the batch cache. If not tagging the batch cache sometimes is ok, then this can be changed separately.)

What is the expected impact of this change? Could it happen that now the ID_RECALC_GEOMETRY tag is missing in many cases? (I don't know what this is doing exactly)

Is this indirectly checking whether there are any modifiers that change the topology?
If yes, then this sounds a bit weak. It can happen that depending on vertex positions a modifier only deforms a mesh or changes the topology (especially with geometry nodes where everything can happen). It feels like checking if the topology has changed can really only be done after modifiers are applied (I'm not sure, maybe modifiers set the ID_RECALC_GEOMETRY flag when necessary. In that case, this concerrn might not apply.)

Generally, a comment for why you are doing these specific checks and what assumptions you make would be good. It is not entirely clear to me why transform code would ever tag with something else than ID_RECALC_GEOMETRY_DEFORM, since it always only deforms.