Determining the formula of zirconium oxide
First, let's solve the problem.
Both relative atomic weight $A_\mathrm{r}$ and relative molecular weight $M_\mathrm{r}$ are a historical dimensionless terms denoting relative atomic mass and molecular mass, respectively:

$$A_\mathrm{r} = \frac{m_\mathrm{a}}{m_\mathrm{u}}$$ 

$$M_\mathrm{r} = \frac{M_\mathrm{B}}{M_\mathrm{u}}$$ 

where $m$ corresponds to atomic mass, $M$ – to molecular mass, and index $\mathrm{u}$ refers to unified atomic mass unit ($= m_\mathrm{a}(\ce{^{12}C})/12$).
Since $\ce{Cl}$ in chlorides is monovalent, then formula of unknown zirconium chloride is $\ce{ZrCl_x}$, where $x$ coincides with the valency of $\ce{Zr}$.
From this the relative weight of $\ce{ZrCl_x}$ is

$$M_\mathrm{r}(\ce{\ce{ZrCl_x}}) = A_\mathrm{r}(\ce{Zr}) + xA_\mathrm{r}(\ce{Cl})$$

relative atomic masses of both elements were known ($A_\mathrm{r}(\ce{Zr})\approx 91$, $A_\mathrm{r}(\ce{Cl})\approx 36$), so that the valency of zirconium can be found via $x$:

$$91 + 36x = 233 \implies x \approx 4$$

Since oxygen in oxides has valency of $2$, then the formula of unknown zirconium oxide is $\ce{ZrO2}$.


  By 'volatile chloride', are they referring to zirconium oxide or something else?


No, they refer to zirconium chloride $\ce{ZrCl4}$.
They mention volatility probably to underline how zirconium and hafnium are separated — by differences in sublimation of their volatile compounds. 


  Also, by the statement 'assuming that zirconium in the oxide has the same valency' are they saying that zirconium has the same valency as chlorine (i.e. -1)?


No, this means that zirconium in both oxide $\ce{ZrO2}$ and chloride $\ce{ZrCl4}$ has the same valency.