I think your first question is, why is there no continuous logarithm function defined in the complex plane, omitting $0$. The answer is, trace out a circle of radius $1$ centered at the origin, starting at $1$, and evaluate the logarithm function as you go around. Go ahead, do it! You will find that you either make a discontinuity somewhere along the way, or else when you get back to $z=1$ you are off by $2\pi i$ from the value you started with.
Now the same thing happens with the argument function. Start it with, say, the value $0$ at $z=1$, go around that circle (counterclockwise, just to be specific), and when you come back to $z=1$ the argument will be $2\pi$; discontinuity! The cure for the discontinuity is to make it impossible to go all the way around the circle. The easiest way to do that is to remove some ray that starts at the origin and goes to infinity. If the ray you remove is the non-positive reals, then the argument, starting with the value $0$ at $z=1$, will increase almost but not quite to $\pi$ as you go around counterclockwise (not reaching the excised ray), and will decrease almost but not quite to $-\pi$ as you go around the other way.
The non-positive imaginary part ray has argument $3\pi/2$ as you approach it counterclockwise from $z=1$, and argument $-\pi/2$ as you approach it the other way, so if you omit that ray then the argument can be taken to be between $-\pi/2$ and $3\pi/2$.
If I have missed the point of your question, please try again.
A point in the UHP is $z=r e^{i\theta}$ with $r > 0$, $0 < \theta < \Pi$. $w = \text{Log}(z)$ maps this to $w = \log(r) + i \theta = u + i v$ with $u = \log(r) \in (-\infty, \infty)$, $v = \theta \in (0,\pi)$. Note that the logarithm of a positive number could be negative. So you have the infinite strip, not a semi-infinite strip.
EDIT:
Hint: half-plane $\implies$ quarter-plane $\implies$ half-disk $\implies$ semi-infinite strip.
Best Answer
One can estimate the Taylor series $$ \operatorname{Log}(1+w) = w - \frac 12 w^2 + \frac 13 w^3 - \frac 14 w^4 + \cdots $$ which converges for $|w| < 1$.
For $|w| \le \frac 12$ is $$ \begin{align} | \operatorname{Log}(1+w)| &\ge |w| - \frac 12 |w|^2 - \frac 13 |w|^3 - \frac 14 |w|^4 - \cdots \\ &\ge |w| - \frac 12 |w| \cdot \underbrace{\left( |w| + |w^2|+ |w|^3 + \cdots\right)}_{\le 1} \\ &\ge |w| - \frac 12 |w| = \frac 12 |w| \, . \end{align} $$
Remark: An upper bound can be obtained in the same way, so that $$ \frac 12 |w| \le \operatorname{Log}(1+w)| \le \frac 32 |w| $$ for $|w| \le \frac 12$.