The delocalised nature of hyperconjugation
I'll try to answer your question through the exemples you give us.
Where are the electrons ?
If you look at the hybridized orbitals making the C-H and the C-C bonds, you can see that the more the bond is polarized, the worse electron donor you have. It's quite understandable since if the bond is polarized, the electron density will be closer to the heteroatom but further to the carbocation that needs stabilization.

Usually the donor ability goes in this order increasingly:
C-F &lt; C-O &lt; C-N &lt; C-H &lt; C-C
Hyperconjugation from C-C to carbocation
Following the &quot;rule&quot; stated before, the stabilization of a carbocation will be best with an adjacent C-C bond. And this is in accordence with what we learn in school since the more alkyl groups nearby, the more stable is the carbacation. Not only because of the increased overall electron density (donor inductive effect) but also because of the hyperconjugation. This is also the base of some reactivity (1,2-C-C bond migration)
Hyperconjugation in the bromonium case
We can easely extend the thing we just saw for the exemple of a bromonium intermediate. In fact it is important to dissociate the cases of the bromonium and the transition state for the bromonium openning.
The bromonium has two bonds with a carbon atom. One could expect the opening reaction to occur in competition between these two sites. We can see that the C-Br bond (Scheme 2, left) can be aligned with a C-C bond, so if we take a look at the molecular orbitals, we will see that this filled sigma C-C bond can interact with the empty C-Br sigma*. Thus creating a destabilization of the C-Br bond (filling a anti-bonding molecular orbital destabilizes the bond) trough an hyperconjugation intercation. But if we look at the second C-Br bond (right) we see that there are no such interaction since there are no filled molecular orbital which can interact with the sigma* of the C-Br bond.

In term of reactivity, the destabilization of the left-hand C-Br bond will favor the breaking of this  bond instead of the other one.