The argument here is not correct because the function $ x \mapsto \frac{2}{3\sqrt{x}}$ is not bounded on $(0,1]$ and is not Riemann integrable. It is true that the improper integral converges, i.e.,
$$\lim_{a \to 0} \int_a^1 \frac{3}{2\sqrt{x}} \,dx = \lim_{a \to 0} (3 - 3\sqrt{a}) = 3,$$
but, in general, Riemann sums of improper integrals may not converge nor be bounded for arbitrarily chosen intermediate points. In fact, for this function the problem will arise unless $c_1$ is specially chosen away from $0$.
As discussed here right- or left-hand sums of monotone functions may converge as the partition is refined but this still leaves a hole in your argument since $c_1 \in (0,x_1)$ is uncontrolled.
(As an aside the part $\sum_{i=1}^n 2c_i(x_i - x_{i-1})$ will converge as $x \mapsto 2x$ is Riemann integrable.)
It appears that you are trying to show that $f(x) = x^2 \sin x^{-3/2}, f(0) = 0$ is of bounded variation on $[0,1]$ by applying the mean value theorem and bounding the derivative as
$$|f'(c_i)| = \left|2c_i \sin c_i^{-3/2}- \frac{3}{2\sqrt{c_i}}\cos c_i^{-3/2} \right| \leqslant 2c_i + \frac{3}{2\sqrt{c_i}}.$$
A valid approach is to write for any partition $P = \{0 = x_0 <x_1< \ldots< x_n = 1\}$,
$$\begin{align}\sum_{i=1}^n |f(x_i) - f(x_{i-1})| &= \sum_{i=2}^n\left|\int_{x_{i-1}}^{x_i}f'(t)\,dt\right| + |f(x_1) - f(0)| \\ &\leqslant \sum_{i=2}^n\int_{x_{i-1}}^{x_i}|f'(t)|\,dt + |f(x_1) - f(0)| \\ &= \int_{x_1}^1|f'(t)| \,dt + |f(x_1)| \\ &\leqslant \int_0^1 \left(2t + \frac{3}{2\sqrt{t}} \right) \, dt + \|f\|_\infty\end{align},$$
where the right hand side is bounded since the improper integral is finite and $f$ is bounded on $[0,1]$. It then follows that $V_f([0,1]) < \infty$.
