Since the resonant structure with a +1 charge on the carbonyl carbon in the ketone is a greater contributor to the overall average structure than the similar carboxylic acid contributor, the carbonyl carbon has a more permanent positive charge and is thus more electrophilic. The partial positive charge is even greater in aldehydes, which only have one alkyl substituent (as opposed to a ketone's two) donating electron density through inductance.
In most reactions, such as nucleophilic substitutions, the leaving group needs to be good (a weak base) in order for the reaction to occur. Thus, we fallaciously associate the carboxylic acid derivatives with the best, most stable leaving groups (acyl halides > anhydrides >> esters ≈ acids >> amides) as the most reactive, when in fact, their carbonyl carbons may not have the most partial positive, and therefore electrophilic, character.
So, NaBH4 can reduce only aldehydes, ketones, and acyl halides, as you said, due to the greater localization of positive charge on the carbonyl carbon than in esters and carboxylic acids, which must be reduced with the more powerful LiAlH4.
Good luck! ;)