By a "continuous choice of $\log f(z)$ along a closed curve $\gamma\colon [a,b]\rightarrow \mathbb{C}\setminus \{0\}$, $\gamma(a)=\gamma(b)$" I suppose you meant a continuous function
$$h\colon \gamma([a,b])\rightarrow \mathbb{C}$$ such that $e^{h(z)}= f(z)$ for every $z\in \gamma([a,b]).$
The winding number of $f$ along $\gamma$ is the winding number of $f\circ \gamma$ around $0$. One way to define the winding number around $0$ is: if $\gamma\colon [a,b]\rightarrow \mathbb{C}\setminus \{0\}$ is continous there is a continuous function (a continuous choice of the argument) $\theta\colon [a,b]\rightarrow \mathbb{R}$ such that
$$f\circ \gamma(t)= |f\circ \gamma(t)|e^{i\theta(t)}$$
for every $t\in [a,b]$. If $\gamma $ is closed, that is $\gamma(a)=\gamma(b)$, then $\theta(b)-\theta(a)= 2\pi k$, with $k\in \mathbb{Z}$. $k$ does not depend on the choice of the function $\theta$ and it is called the winding number of $f\circ \gamma$. It follows that
$$\log f(\gamma(t)):= \log |f(\gamma(t))| + i \theta(t).$$
is continuous on $[a,b]$. But note that $\log f(\gamma(a))= \log f(\gamma(b))$ if and only if the winding number of $f\circ \gamma$ is $0$.
(1) Suppose that $f\circ \gamma$ is simple (non self-intersections) with winding number $0$ around $0$. Then there is a continuous choice of the logarithm of $\log f(z)$ along $\gamma$.
Indeed if the winding number is $0$ then $\theta(a)=\theta(b)$. If $\gamma$ is simple (non self-intersections) then for $z=\gamma(t)$ we define $\log f(z)=\log f(\gamma(t))$. This is well defined and continuous since $\gamma$ is simple and $\theta(a)=\theta(b)$. So $\log f(z)$ is a continuous choice of the logarithm on $\gamma([a,b])$.
Note that $(1)$ does not hold if we remove the assumption that $f\circ \gamma$ is simple. Indeed consider the curve $\gamma$ as below and $f(z)=z$. $f\circ \gamma= \gamma$ has winding number $0$ but there is not way to define a continuous choice of $\log f$ on the image of the curve $\gamma$, since it contains a closed subcurve that turns once around $0$ and pass through $C$. When you move along this curve once you need to subtract $2\pi$ from the imaginary part of $\log z$.
On the other hand if there is a continuous choice of $\log f(z)$ along a closed curve $\gamma$ then the winding number of $f\circ \gamma$ is zero since we can take $\theta(t)= (Im (\log f))(\gamma(t))$ as a continuous choice of the argument and this implies $\theta(a)=\theta(b)$.
