R & S Stereochemistry of an unsaturated ring system
I don’t really know how you arrived at your reasoning, but it seems to fail as soon as double bonds are involved. If you encounter a double bond, it must be expanded with ghost atoms to two single bonds in the following manner:

$$\ce{(R2)C=C(R'2) -&gt; (R2)C(C)-C(C)(R'2)}$$

The carbon atoms in bracket are ghost atoms not connected to anything else. These ghost atoms are treated just like normal atoms when determining priorities. You can read more on Wikipedia, which also includes a better graphic example. (But since your case is rather simple, we can do without.)

Let’s first determine the tertiary alcohol’s carbon’s absolute configuration:


$\ce{O}$ from the hydroxy group.
$\ce{C}$ (along the ring to the left)


$\ce{C}$ (the quarternary carbon)


$\ce{C}$
$\ce{C}$ ← This is the first point of difference.
$\ce{(C)}$ — this is the double bond’s ghost atom.

$\ce{H}$
$\ce{H}$

$\ce{C}$ (along the ring upwards)


$\ce{C}$


$\ce{C}$
$\ce{H}$ ← This is the first point of difference.
$\ce{H}$

$\ce{H}$
$\ce{H}$

$\ce{C}$ from the methyl group.


$\ce{H}$
$\ce{H}$
$\ce{H}$



Thus, the priorities are $\ce{OH} &gt; \ce{CH2C_{(q)}} &gt; \ce{CH2CH2R} &gt; \ce{CH3}$, because the quarternary carbon that one of the two $\ce{CH2}$ groups is bonded to beats the secondary carbon at the top of the right-hand ring. That makes that carbon’s absolute configuration (S).

Similarly, for the carbon attached to iso-propyl:


$\ce{C}$ (the quarternary carbon)


$\ce{C}$
$\ce{C}$
$\ce{C}$

$\ce{C}$ (the iso-propyl group)


$\ce{C}$
$\ce{C}$
$\ce{H}$

$\ce{C}$ (the secondary carbon)


$\ce{C}$
$\ce{H}$
$\ce{H}$

$\ce{H}$


As you can see, we only need go down to the second level to find a difference: $\ce{C(C3)} &gt; \ce{C(C2H)} &gt; \ce{C(CH2)} &gt; \ce{H}$ Thus, that carbon’s absolute configuration is (R).