The argument-principle integral is the winding number of the image cycle

Since $f$ is continuous on the compact set ${\gamma }^{\ast }$ and never vanishes there, $f\circ \gamma$ is a closed complex contour whose trace misses $0$. By [L3], choose a continuous logarithm $\lambda$ of $f\circ \gamma$. Cover ${\gamma }^{\ast }$ by finitely many open discs $U_1,\dots ,U_N$ on which $f$ has no zeros, and then subdivide $\gamma$ into consecutive subcontours ${\gamma }_j$ whose traces lie in those discs.

Fix $j$. On $U_j$, [L4] gives a holomorphic logarithm $L_j$ of $f$, with $L_j^{\mathrm{\prime }}=f^{\mathrm{\prime }}\mathrm{/}f$. Along the trace of ${\gamma }_j$, both $L_j\circ {\gamma }_j$ and $\lambda {\mathrm{\mid }}_{{\gamma }_j}$ are continuous logarithms of $f\circ {\gamma }_j$, so [L3] makes their difference constant. Therefore $\lambda (t_j)-\lambda (t_{j-1})=L_j(\gamma (t_j))-L_j(\gamma (t_{j-1}))={\int }_{{\gamma }_j}\frac{f^{\mathrm{\prime }}(z)}{f(z)}dz,$ where the last equality is [L5] applied to the primitive $L_j$.

Summing the equalities of step 2.1 over the subdivision and using the additivity from [L5] gives ${\int }_{\gamma }\frac{f^{\mathrm{\prime }}(z)}{f(z)}dz=\lambda (b)-\lambda (a)\mathrm{.}$ Now [L1] and [L2] applied to the contour $f\circ \gamma$ identify the same increment with both $2\pi in(f\circ \gamma ,0)$ and $i(\theta (b)-\theta (a))$, so the two displayed formulas follow.