I cannot comprehend $$\int_a^bf(\gamma(t))d|\gamma|(t).$$ Context (brackets mine):
Let $\gamma:[a,b]\to\mathbb C$ be a rectifiable path and for $a\leq t\leq b$, let $|\gamma|(t)$ be $V(\gamma;[a,t])$ (in words, I think this means something like "the total variation of $\gamma$ restricted to $[a,t]$"). That is, $$ |\gamma|(t)=\sup\left\{\left.\sum_{k=1}^n|\gamma(t_k)-\gamma(t_{k-1})|\ \right| \{t_0, \dots, t_n\} \text{ is a partition of } [a, t]\right\}. $$ Clearly $|\gamma|(t)$ is increasing and so $|\gamma|:[a,b]\to\mathbb R$ is of bounded variation. If $f$ is continuous on $\{\gamma\}$ (the trace of $\gamma$) define $$ \int_{\gamma}f|dz|=\int_a^bf(\gamma(t))d|\gamma|(t). $$ Conway, JB. 1978. Functions of one complex variable I. (2nd ed.). New York: Springer. p. 64.
Which leads to my question, about the last integral expression. We are integrating with respect to $|\gamma|(t)$, a function, but I do not understand how one is to do that when the function $|\gamma|(t)$ does not feature in the integrand $f(\gamma(t))$. Basically, it went over my head.
You can think of it like this: fix $t$ then calculate $y = \gamma(t)$ and $x = |\gamma|(t)$. Since $x$ is monotonic you can find its inverse, so if you are given $x$ you can calculate $t$ and then $y$. Say in another words given $x$ you can always find $y(x)$. The integral is then
$$ \int dx\;f(y(x)) $$
which I guess will make more sense to you