The Argument Principle says that for a meromorphic function $f(z)$, and a closed contour $C$, $\int_{C} \frac{f'(z)}{f(z)} \, dz=2 \pi i(Z-P)$, with $Z$ the number of roots in $C$, and $P$ the number of poles in $C$. Assume $C$ is written in polar form as $C_p(\theta)$, $0 \le \theta \le 2 \pi$, so the integral at this point is $= \int_{0}^{2 \pi} \frac{f'(C_p(\theta))}{f(C_p(\theta))} \, C_p'(\theta) \, d\theta$.
Due to the fact that $\int \frac{f'(z)}{f(z)} dz=\ln(f(z))$ and using the rule of integration by substitution, is it true that the integral can be reduced to $\ln(f(C_p(2 \pi)))-\ln(f(C_p(0)))$ (+ some constant)?
A comment that got too long:
I think that you are missing the essence of the argument principle here - for any continuous function $f$ that is (finite and) non zero on a simple curve $\gamma$, one can define a continuous logarithm of $f$ on the curve with the usual properties so for example if the curve is closed and parametrized say by $\gamma(t), t \in [0,1]$, $f(\gamma (t))=e^{\alpha(t)}, \alpha$ continuous on $[0,1]$ and $\alpha(1)-\alpha(0)=2\pi in(f)$ for some integer (called the degree of $f$ wr $\gamma$ or equivalently $2\pi in$ is called the argument change of $f$ when it goes around $\gamma$) that depends only on $f$ but not on the parametrization of the curve.
In particular the integral in the OP can always be computed as a difference of logarithms when the function $f$ is just real differentiable on $\gamma$ rectifiable (so in other words we need only $t \to f(\gamma(t))$ differentiable in $t \in [0,1]$ and some similar smoothness condition on $\gamma$) as then we can construct $\alpha$ differentiable and $f'd\gamma/f=\alpha'(t)dt$
The point of the argument principle in the usual complex analysis setting of holomorphic/meromorphic functions is that there is a simple way to compute $n(f)$ in terms of the orders of zeroes and poles of $f$ inside $\gamma$ as long as $\gamma$ is a rectifiable Jordan curve and $f$ is meromorphic inside $\gamma$ and continuous non-zero finite on $\gamma$; this seems "obvious" but simple examples of harmonic (hence quite smooth) functions with for example a simple zero at the origin in the unit disc and no other zeroes, but of zero degree, so the integral is zero shows the power of the classical result
(eg $2\Re z+iz^2$ is harmonic and has an isolated zero at the origin of "multiplicity" zero and no other zeroes inside the closed unit disc, so its argument doesn't change when going around by 360 degrees on the unit circle as one can show that $2\Re z+iz^2$ avoids the positive imaginary axis when $|z|<1$)