Hello, fellow o-chemer!
So you would think that a carboxylic acid would be more electrophilic than, say, a ketone because there is more than one oxygen attached to the carbonyl carbon to withdraw electron density inductively, right?
Well, usually it isn't quite that simple and there are more things going on. True, both oxygen atoms inductively withdraw electron density, but they also both donate it by resonance as well. Let's say that our carbonyl oxygen is feeling greedy and pulls the pi-bond electrons up to be its own lone pairs (as happens in an important resonance contributor) to gain a -1 charge, and forcing the carbon to gain a +1 charge. Immediately, the other oxygen of the carboxilic acid's -OH group can donate a lone pair of its own to gain an unfavorable, but possible, +1 charge and restore the carbon's neutrality. These are all resonance structures that contribute to the overall average charge on each of the atoms involved.
Now, let's say that a ketone's carbonyl oxygen sees this and decides, "hey, I want to be greedy too!" as it takes the pi bonded elections to become its lone pair and gains a charge as -1 and forces the carbonyl carbon to adopt a +1 charge, as happened before. Now, there is no other oxygen with a lone pair to come to the rescue with another lone pair to collapse down into a pi bond to the carbonyl carbon, so the carbon is stuck with a +1 charge.
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! ;)
UC Santa Barbara edumacation!