A TA talked about this today in the context of ketones.
If you stick an acetone and $\ce{NaBH4}$ together in a protic solvent, you get 2-propanol.
If you stick acetone and $\ce{NaH}$ together, you get an enolate. What's the difference? They're both hydrides.
First of all, borohydride as a group is not same as hydride. So the mechanisms are very different.
Because a simple $\ce{H-}$ anion is more basic than an enolate $\ce{C=C-O^{-}}$(product of $\alpha$ hydrogen deprotonation of a ketone), it just deprotonate before reduction will happen. Deprotonated ketones (enolates) will no longer be a substrate for reduction.
Borohydride $\ce{BH4-}$ is not as basic as enolate, so deprotonation hardly happen. Thus a reduction will be the predominant reaction.
To complement the answer provided by @Ian Fang and edited by @santiago, sodium hydride's action as a Brønsted base differs from other bases.
"Normal" acid base reactions are reactions of an equilibrium, like the partial deprotonation of malonic esters by 5 m% sodium carbonate in some Knoevenagel reactions:
$\ce{R-H} + \mbox{base} \rightleftharpoons{} \ce{R-} + \mbox{protonated base}$
Deploying sodium hydride as base however often intends a quantitative deprotonation. The equilibrium is driven to the right hand side, in favour of the protonated base ($\ce{H2}$) immediately leaving the reaction mixture -- so the back reaction can't take place.
