I have arrived at the inequality
$$ V(\gamma) \le \sup\limits_{P\in\mathcal{P}[a,b]}\sum\limits_{i=1}^n\int\limits_{t_{i-1}}^{t_i} \left| \gamma'(s)\right|ds $$
where $\gamma(t):[a,b]\to\mathbb{C}$ is a smooth parametric curve, which is a uniformly continuous function, $V(\gamma)$ is the total variation of $\gamma$, and $P\in\mathcal{P}[a,b]$ means a partition $P$ $(a=t_0<t_1...<t_n=b)$ in the set of all partitions on $[a,b]$.
Since $\gamma$ is uniformly continuous, it must be possible to interchange the summation and the integration signs. But then the limits of integration must also interchange. But how? And what to do with the supremum, should it also be taken under the integral sign?
Integrals are additive with respect to the interval of integration. For example, $$ \int_1^3 f(x)\,dx + \int_3^4 f(x)\,dx = \int_1^4 f(x)\,dx $$ If the above its understood, it should become clear that for any partition of the interval $[a,b]$ the sum of integrals over the subintervals of the partition is equal to the integral over $[a,b]$. Hence,
$$V(\gamma) \le \sup\limits_{P\in\mathcal{P}[a,b]}\int_a^b \left| \gamma'(s)\right|ds$$
And it should be more clear what to do with the supremum now. No, don't rearrange symbols in a random pattern just for the sake of doing something. Think about what the supremum is taken over: it's over partitions $P$. But the integral does not involve the partition in any way. So it's the supremum of a one-element set, which means it is equal to that element, $$\int_a^b \left| \gamma'(s)\right|ds$$