Here $\nabla$ is the levi-civita connection of the given metric $g$. I am stuck at the last equality (g). What on earth does $\nabla^a \nabla_a$ mean? Isn't it just $g^{ab}\nabla_a \nabla_b$? But, then the equality does not hold because the left side of (g) is just $g^{ab}\partial_a\partial_bf$ and the right side of (g) contains more than that. Could anyone please help me what $\nabla^a \nabla_a$ mean?
You are right for the first statement, i.e., $$ \nabla^a\nabla_af=g^{ab}\nabla_a\nabla_bf. $$ However, your second statement remains to be improved. Provided that $$ \nabla_bf=\partial_bf $$ is a vector, $$ \nabla_a\nabla_bf=\nabla_a\left(\partial_bf\right)=\partial_a\left(\partial_bf\right)-\Gamma_{ab}^c\left(\partial_cf\right). $$ Therefore, $$ \nabla^a\nabla_af=g^{ab}\nabla_a\nabla_bf=g^{ab}\left(\partial_a\partial_bf-\Gamma_{ab}^c\partial_cf\right). $$ This is still faraway from the ultimate goal. To make it short, we could turn to part (e) for help.
Provided that, using $\nabla_ag^{bc}=0$, $$ \nabla^a\nabla_af=g^{ab}\nabla_a\nabla_bf=\nabla_a\left(g^{ab}\nabla_bf\right)=\nabla_a\left(g^{ab}\partial_bf\right) $$ and that, from part (e), $$ \nabla_aV^a=\left|g\right|^{-1/2}\partial_a\left(\left|g\right|^{1/2}V^a\right), $$ let $V^a=g^{ab}\partial_bf$, and the last formula immediately gives $$ \nabla^a\nabla_af=\nabla_a\left(g^{ab}\partial_bf\right)=\left|g\right|^{-1/2}\partial_a\left(\left|g\right|^{1/2}g^{ab}\partial_bf\right), $$ as is expected.