First, we need to draw in all the lone pairs.  Once we have done that, we should work out which bonds have been made and broken.

The curly arrows must account for ALL of these bonding changes. Since we are looking at an S_N2 process, the Nu attacks as the LG is displaced:

Notice that these 2 curly arrows only show the formation of the C-O bond and the breaking of the C-Br bond and that the CHARGES BALANCE (2 neutral starting material molecules give products that are a negative and a positive ion). Since the neutral O in water gave away electrons to form the new C-O bond, the O becomes positive in the intermediate ion (an oxonium ion).

To complete the reaction we need to show the loss of a proton breaking the O- bond and the formation of the H-Br bond.

Do the charges balance ?
Yes, we have +ve and -ve starting materials whose charges cancel going to a neutral product molecule.

Now for the reverse reaction, i.e. the reaction of isopropanol with HBr to give isopropyl bromide.

This reaction must occur via the same pathway as we used above (the Law of Microscopic Reversibility).  However, we should be able to deduce what happens without knowing that !
First, we know that HBr is a strong acid and therefore we should expect it to protonate a base (after all, that's what acids do !).
The most basic sites in the whole system are the lone pairs on the O. Since the lone pairs are the electron rich center, that arrows must start there and go towards the proton. This is a simple acid-base reaction.