You are not alone, many people are confused about the moment of inertia and polar moment of inertia of the circular shape, $I, I_p$, and their uses.
Let's review two equations:
$\sigma_b = \dfrac {My}{I}$, where $\sigma_b$ is the normal stress induced by the bending moment $M$, and $I$ is the moment of inertia. For circle, $I = \dfrac {\pi d^4}{64}$, where $d$ is the diameter of the circle.
$\tau = \dfrac {TL}{J}$, where $\tau$ is the shear stress induced by the torsional force $T$, and $J$ is the polar moment of inertia, sometimes written as $I_p$. For circular shape, $J$ or $I_p = \dfrac {\pi d^4}{32}$
The cantilever beam in the example subjects to combined stresses of "tension", "bending", "shear" and "torsion". Therefore,
$\sigma_{A,B} = \dfrac {F_t}{A} \pm \dfrac {My}{I_x}$ - $Normal Stress$
$\tau_{A,B} = \sqrt (F_y/A)^2 + (TL/I_p)^2$ - $Shear Stress$
Note: The concept above ($"M" vs "I"$; $"T" vs "I_p"$) applies to all shapes but differs in the form of the polar moment of inertia ($I_p$ or $J$).
For a circular shaft, the maximum and minimum stresses always fall on the perimeter at the opposite points (extreme points). The locations depend on the direction of the bending moment and the nature of the axial force, whether it is a tensile or compressive force. You can stop at finding and expressing the shear stress and normal stress separately for most of the practical concerns, but occasionally (especially for academic works), you might need to consider the maximum point stress by combining the shear stress and normal stress using the method of "square root of squares, $c = \sqrt{a^2 + b^2}$).
Note, the maximum/minimum stress locations stated above do not hold for other shapes.