Let's agree explicitly that "$\log$" refers to the branch of logarithm defined on $\mathbf{C}\setminus(-\infty, 0]$ whose imaginary part is between $-\pi i$ and $\pi i$, and that "$f(z) \leq a$" means "both $a$ and $f(z)$ are real, and $f(z) \leq a$".
If $f$ is holomorphic in some region $U$, then $\log(f)$ is holomorphic at $z$ in $U$ provided $f(z)$ does not lie on the branch cut, i.e., "$f(z) \leq 0$" is false.
In your example, $f(z) = z^{2} + a^{2}$ is holomorphic on all of $\mathbf{C}$, so $\log f$ is holomorphic on the twice-cut plane, from which the imaginary intervals $i(-\infty, -a]$ and $i[a, \infty)$ have been removed. These intervals are precisely the set of $z$ for which $z^{2} + a^{2} \leq 0$. In fancier language, they constitute the preimage of the cut $(-\infty, 0]$ under $f$.
The issue in (2) is that a small circle $C$ centered at $ib$ (with $|b| > a$) crosses the "cuts" of $\log(f)$. The image of $C$ under $f$ is a closed curve (not a circle) "centered on" the negative real axis, so $f(C)$ crosses the branch cut of $\log$. Locating $-b^{2} + a^{2}$ in the $z$ plane, relative to the branch cuts of $\log(f)$, is immaterial: We're not evaluating $\log(f)$ near $-b^{2} + a^{2}$, we're evaluating $\log$ itself. $\ddot\smile$
To see what's going on with the values of $\log(f)$ "near" each imaginary cut, consider points $z = re^{i\phi}$ with $r > a$ and $\phi$ "close to" $\pm\pi/2$:
$$
z^{2} + a^{2}
= r^{2} e^{2i\phi} + a^{2}
= (r^{2} \cos(2\phi) + a^{2}) + ir^{2} \sin(2\phi).
$$
If $\phi$ is a bit smaller than $\pm\pi/2$, then $\sin(2\phi) > 0$. Consequently, $z^{2} + a^{2}$ has "small" positive imaginary part, so $\log(z^{2} + a^{2})$ has imaginary part close to $\pi i$.
Similarly, if $\phi$ is a bit larger than $\pm\pi/2$, then $\sin(2\phi) < 0$ and $\log(z^{2} + a^{2})$ has imaginary part close to $-\pi i$.
There is no difference between your approaches. Remember that the exponential function is periodic with period $2\pi i$. Therefore, $e^{-2+3i+2i\pi k}$ is always the same number, no matter the choice of $k\in\mathbb Z$.
Best Answer
Keep always in mind that the logarithm of $y$ is just the number $x$ such that $e^x=y$
On the real numberline, this is quite straightforward. If $y$ is positive, there is a unique $x \in \mathbb{R}$ such that $e^x=y$, therefore $log(y)=x$ If y is zero or negative, there is no such number $x$, thus $log(y)$ in undefined.
On the complex plane, however, for any given $z \neq 0$, there are infinitely many complex numbers $w$ that satisfy $e^w=z$, since the expression $\log z = \log r + i\theta$ allows for \theta to increase in multiples of $2\pi$ with $z$ remaining the same. The multiple branches of the logarithm come from here
In short, the complex logarithm is not really a function in the sense we often think of functions as mappings that assign one unique value to each member of the domain set. It is a function if we restrict ourselves to bands of the complex plane with height $2\pi$