Branch cut for $\log (-z)$

You would like to describe $\log$ as the inverse function of $e$. So you would like to say that $\log(re^{i\theta}) = \log r + i \theta$.

But then you have a problem because $\theta$ is only defined up to $2\pi$. Conclusion : every logarithm on $\mathbb C$ has a cut on the argument. But not on the radius, so the cut is always a half-line. This half-line correspond to where you start counting the argument.

For example, in your first example, you start counting at $\rm{arg}$ $(z)= 0$ i.e the half-line $[0, +\infty[$.

In your second example, you start at $\rm{arg}$ $(z) = \pi$, which is the half line $]-\infty, 0]$.

To add to Damien's good answer: in fact, there are further options, which are sometimes useful. Namely, as long as you "cut" along some (not self-intersecting) path that connects the two "bad" points for log, 0 and $\infty$, you can define an unambiguous log on what's left (by $\log(z)=\int_\gamma dz/z$, where $\gamma$ is a path from $1$ to $z$ that does not cross your cut).

And looking at $\log(-z)$ doesn't really change anything: you could use the same old cut, if you wanted, since the goal is to disambiguate (in some way or another, in the usual simple fashion as described by Damien, or, possibly, in a trickier way to suit circumstances).

Edit: added in response to comment... It is not precisely true that $\arg(-z)=\arg(-1)+\arg(z)$ as numbers, because of the ambiguity (by $2\pi\mathbb Z$) in both of these "arg"s. For that matter, do we want $\arg(-z)=\arg(z)+\pi$, or do we want $\arg(z)-\pi$? Cases can be made for both. There's simply no compulsion to choose one or the other.

And, back to the more general point, there is no compulsion to have the "cut" be a straight line from 0 to $\infty$.

An example to illustrate that our "cuts" really have no impact on the mathematical objects themselves: we know that a power series for a holomorphic functions converges up to the first "singularity". Ok. Consider the power series of $\log(z)$ at $-4+3i$. If we believe that somehow the plane is "cut" along the negative real axis, we'd think that the radius of convergence would be $3$. However, it is $5$. The function ignores our "cutting" entirely.