The two concepts match. Let us at first revisit the logarithmic function:
The multivalued logarithm is defined as
\begin{align*}
\log(z)=\log|z|+i\arg(z)+2k\pi i\qquad\qquad k\in\mathbb{Z}\tag{1}
\end{align*}
In order to make single-valued branches of $\log $ we make a branch cut from $0$ to infinity, the most common being the negative real axis. This way we define the single-valued principal branch or principal value of $\log$ denoted with $\mathrm{Log}$ and argument $\mathrm{Arg}$. We obtain
\begin{align*}
\mathrm{Log}(z)=\log |z|+i\mathrm{Arg}(z)\qquad\qquad -\pi <\mathrm{Arg}(z)\leq \pi\tag{2}
\end{align*}
Now let's look at the square root function:
The two-valued square root is defined as
\begin{align*}
z^{\frac{1}{2}}&=|z|^{\frac{1}{2}}e^{i\frac{\arg(z)+2k\pi}{2}}\\
&=|z|^{\frac{1}{2}}e^{i\frac{\arg(z)}{2}}(-1)^k\qquad\qquad k\in\mathbb{Z}\tag{3}
\end{align*}
In order to make single-valued branches of $z^{\frac{1}{2}}$ we make again a branch cut from $0$ to infinity along the negative real axis. This way we define the single-valued principal branch or principal value of $z^{\frac{1}{2}}$ denoted with $\left[z^{\frac{1}{2}}\right]$ and argument $\mathrm{Arg}$. We obtain
\begin{align*}
\left[z^{\frac{1}{2}}\right]&=|z|^{\frac{1}{2}}e^{i\frac{\mathrm{Arg}(z)}{2}}\qquad\qquad -\pi <\mathrm{Arg}(z)\leq \pi\tag{4}
\end{align*}
Now we are ready to calculate $e^{\frac{1}{2}\log(z)}$
We obtain from (1)
\begin{align*}
\color{blue}{e^{\frac{1}{2}\log(z)}}&=e^{\frac{1}{2}\left(\log|z|+i\arg(z)+2k\pi\right)}\\
&=|z|^{\frac{1}{2}}e^{\frac{1}{2}\left(i\arg(z)+2k\pi\right)}\\
&=|z|^{\frac{1}{2}}e^{i\frac{\arg(z)}{2}}(-1)^k\\
&\color{blue}{=z^{\frac{1}{2}}}
\end{align*}
which coincides with (3).
Taking the principal value $\mathrm{Log}$ we obtain from (2)
\begin{align*}
\color{blue}{e^{\frac{1}{2}\mathrm{Log}(z)}}&=e^{\frac{1}{2}\left(\log |z|+i\mathrm{Arg}(z)\right)}\\
&=|z|^{\frac{1}{2}}e^{i\mathrm{Arg}(z)}\\
&\color{blue}{=\left[z^{\frac{1}{2}}\right]}
\end{align*}
which coincides with (4).
We also see the relationship
\begin{align*}
e^{\frac{1}{2}\log(z)}=\left[z^{\frac{1}{2}}\right](-1)^k
\end{align*}
Conclusion: The concepts of logarithm and square root match in the sense that the infinitely many branches of the logarithm yield precisely the two branches of the square root.
Note: This answer is mostly based upon chapter VI from Visual Complex Analysis by T. Needham.
The concept of a branch of the argument function on a domain $U \subset \mathbb C^* = \mathbb C \setminus \{0\}$ should include continuity. If we drop this requirement, we would allow completely erratic argument functions. So the definition should be
A continuous function $A :U \to \mathbb R$ is called a branch of the argument function on $U$ if $\lvert z \rvert e^{iA(z)} = z$ for all $z∈U$.
Given such a branch $A$ we define
$$\ln_A : U \to \mathbb C, \ln_A(z) = \ln^{\mathbb R}(\lvert z \rvert) + i A(z) .$$
Here $\ln^{\mathbb R} : (0,\infty) \to \mathbb R$ denotes the real logarithm. Clearly $\ln_A$ is a continuous function such that $e^{\ln_A(z)} = e^{\ln^{\mathbb R}(\lvert z \rvert)}e^{i A(z)} = \lvert z \rvert e^{i A(z)} = z$ for all $z \in U$. Note that if we take a non-continuous $A$, then also $\ln_A$ is not continuous so that it cannot be holomorphic.
We claim that $\ln_A$ is holomorphic for each branch $A$ of the argument function on $U$. That is, $\ln_A$ is a branch of the logarithm on $U$.
Consider $z_0 \in U$. Since the exponential map has a nowhere vanishing derivative, it maps an open neighborhood $W$ of $\ln_A(z_0)$ biholomorphically onto an open neighborhood $V$ of $e^{\ln_A(z_0)} = z_0$. Let $E : W \to V$ denote the biholomorphic restriction of the exponential map. Since $\ln_A$ is continuous, we find an open neighborhood $V'$ of $z$ in $U$ such that $\ln_A(V') \subset W$. Let $V'' = V \cap V'$. For $z \in V''$ we get $E(\ln_A(z)) = e^{\ln_A(z)} = z = E(E^{-1}(z)$, thus $\ln_A(z) = E^{-1}(z)$. This shows that $\ln_A$ is complex differentiable at $z_0$.
In fact we have a $1$-$1$-correspondence between branches of the argument function on $U$ and branches of the logarithm on $U$. To see this, let $\Re, \Im : \mathbb C \to \mathbb R$ denote the real and imaginary part functions $\Re(x + iy) = x, \Im(x + iy) = y$.
Given a branch $\ell$ of the logarithm on $U$, the function $\Im \circ \ell$ is a branch of the argument function on $U$: We have $z = e^{\ell(z)} = e^{\Re(\ell(z))} e^{i \Im(\ell(z))}$ which implies $\lvert z \rvert = e^{\Re(\ell(z))}$ and thus $z = \lvert z \rvert e^{i \Im(\ell(z))}$.
The associations $A \mapsto \ln_A$ has $\ell \mapsto \Im \circ \ell$ are inverse to each other:
$\ln_{\Im \circ \ell} = \ell$:
Since $\lvert z \rvert = e^{\Re(\ell(z))}$, we get $$\ln_{\Im \circ \ell}(z) = \ln^{\mathbb R}(\lvert z \rvert) + i (\Im \circ \ell)(z) = \ln^{\mathbb R}(e^{\Re(\ell(z))}) + i \Im(\ell(z)) = \Re(\ell(z)) + i \Im(\ell(z)) = \ell(z).$$
$\Im \circ \ln_A = A$:
We have $\Im(\ln_A(z))) = \Im( \ln^{\mathbb R}(\lvert z \rvert) + i A(z)) = A(z)$.
Let us now show that the function $f : D \to \mathbb R$ is well-defined.
The issue here is that the functions $Arg_\alpha$ are not defined on all of $\mathbb C^*$, but only on the sliced planes $S_\alpha = \mathbb C \setminus \{te^{i\alpha} \mid t \in [0,\infty)\}$. Thus we have to verify that $z \in S_{\lvert z \rvert - 2\pi}$ for all $z \in D$. Assume that $z \notin S_{\lvert z \rvert - 2\pi}$. Then $z = te^{i(\lvert z \rvert -2\pi)} = te^{i\lvert z \rvert}$ for some $t \ge 0$. This gives $\lvert z \rvert = t$, thus $z = te^{it} \notin D$, a contradiction.
Let us next show that $(2)$ is correct.
For $z \in D$ we have $z = \lvert z \rvert e^{if(z)}$, where $f(z) \in (\lvert z \rvert - 2\pi, \lvert z \rvert)$. Thus $e^{i(\pi - \lvert z \rvert)} \cdot z = \lvert z \rvert e^{i(\pi - \lvert z \rvert +f(z))}$, where $\pi - \lvert z \rvert + f(z) \in (-\pi,\pi)$. Thus $e^{i(\pi - \lvert z \rvert)} \cdot z$ is contained in the sliced plane on which the principal argument $Arg$ is defined; we have $Arg(e^{i(\pi - \lvert z \rvert)} \cdot z) = \pi - \lvert z \rvert + f(z)$ which is nothing else than $(2)$.
But it is obvious that the RHS of $(2)$ is a continuous function which means that $f$ is continuous.
We have now answered your questions 1. - 3. Question 4. is too vague to be answered.
Update:
Here is an approach avoiding to use the inverse function theorem for complex functions.
Let $A : U \to \mathbb R$ be a branch of the argument function. Then
$$\cos(A(z)) + i \sin(A(z)) = e^{iA(z)} = \frac{z}{\lvert z \rvert} $$
Identifying $\mathbb C$ with $\mathbb R^2$ we get
$$\cos(A(x,y)) = \frac{x}{\sqrt{x^2+y^2}}, \sin(A(x,y)) = \frac{y}{\sqrt{x^2+y^2}} .$$
$A(x,y)$ is contained in one or two one of the intervals of the form $J^c_k = (k\pi, (k+1)\pi)$ or $J^s_k = (k\pi - \frac \pi 2, k\pi + \frac \pi 2)$ with $k \in \mathbb Z$ (it is contained in exactly one of these intervals iff $A(x,y)$ is an integral multiple of $\frac \pi 2$). The intervals $J^c_k$ are mapped by $\cos$ bijectively onto $(-1,1)$, the intervals $J^s_k$ are mapped by $\sin$ bijectively onto $(-1,1)$.
Let us assume that $A(x,y) \in J^c_k$ (the case $A(x,y) \in J^s_k$ can be treated similarly). The inverse function $(-1,1) \to J^c_k$ is a branch of the arcus cosinus which we denote by $\arccos_k$. We know that $\arccos_k$ is continuously differentiable.
By continuity $A$ maps an open neigborhood $V \subset U$ of $(x,y)$ into $J^c_k$. Thus on $V$ we have
$$A(x,y) = \arccos_k\left(\frac{x}{\sqrt{x^2+y^2}}\right). \tag{1}$$
This shows that $A$ is continuously differentiable on $V$.
Since this is true for all $(x,y) \in U$, we see that $A$ is continuously differentiable on $U$.
This shows that $\ln_A(z) = \ln^{\mathbb R}(\lvert z \rvert) + i(A(z)$ is real differentiable.
We can now compute the partial derivatives of $u(x,y) = \ln^{\mathbb R}(\sqrt{x^2+y^2})$ and $v(x,y)= A(x,y)$. Doing this for $A$ requires to treat of a lot of cases (we have to treat $(1)$ and its $\sin$-version, and in the derivatives we have to be careful with signs). Anyway, we should be able to verify the Cauchy–Riemann equations which assure complex differentiability. I have not done it, it is tedious work.
Another proof of complex differentiability without verifying the Cauchy–Riemann equations is based on A simple proof that a real differentiable local section of a holomorphic function is holomorphic.
Best Answer
Using the definition of the complex exponential, we have
$\sqrt{z(z-1)}=e^{\frac12\log(z(z-1))}$
In term of set equivalence, $\log(z(z-1))=\log(z)+\log(z-1)$. This means that any value of $\log(z(z-1))$ can be expressed as some value of $\log(z)$ and some value of $\log(z-1)$. Conversely, it means that the some of any value of $\log(z)$ and any value of $\log(z-1)$ can be expressed as some value of $\log(z(z-1))$. Therefore, we can write (as set equivalence)
$$\sqrt{z(z-1)}=\sqrt{z}\sqrt{z-1}$$
There are branch points at $z=0$ and $z=1$ (and one at the point at infinity). We are free to choose branch cuts that begin at these branch points and extend to the point at infinity.
Let's choose branch cuts from $z=0$ and $z=1$ that are rays along the negative real axis such that
$$-\pi\le\arg(z)<\pi\\$$ $$\pi\le\arg(z-1)<3\pi$$
At $z=2$, $\sqrt{z}=\sqrt{2}$ and $\sqrt{z-1}=-1$ so that $\sqrt{z(z-1)}$
We could have chosen instead the branches in which
$$\pi\le\arg(z)<3\pi\\$$ $$-\pi\le\arg(z-1)<\pi$$
Then, we see that at $z=2$, we have $\sqrt{z}=-\sqrt{2}$ and $\sqrt{z-1}=1$ so that $\sqrt{z(z-1)}=-\sqrt{2}$ as expected!