Hmm. It doesn't seem to be a very big gap.
If [itex]Q[/itex] is a function of coordinates, then [itex]\dfrac{d}{d\tau} Q = \dfrac{dx^{\gamma}}{d\tau} \nabla_{\gamma} Q[/itex]. So in the particular case [itex]Q = w_{\alpha \beta} \xi^{\alpha}\eta^{\beta}[/itex], we use the product rule to get
[itex]\dfrac{d}{d\tau}(w_{\alpha \beta} \xi^{\alpha}\eta^{\beta}) = <br />
((\nabla_{\gamma} w_{\alpha \beta}) \xi^{\alpha}\eta^{\beta}<br />
+ w_{\alpha \beta} (\nabla_{\gamma} \xi^{\alpha}) \eta^{\beta}<br />
+ w_{\alpha \beta} \xi^{\alpha} (\nabla_{\gamma} \eta^{\beta})) \dfrac{dx^{\gamma}}{d\tau}[/itex]
Using the semicolon notation, and using [itex]\dfrac{dx^{\gamma}}{d\tau}= u^{\gamma}[/itex], this becomes:
[itex]\dfrac{d}{d\tau}(w_{\alpha \beta} \xi^{\alpha}\eta^{\beta}) = <br />
(w_{\alpha \beta ; \gamma}\ \xi^{\alpha}\ \eta^{\beta}<br />
+ w_{\alpha \beta}\ \xi^{\alpha}_{; \gamma}\ \eta^{\beta}<br />
+ w_{\alpha \beta}\ \xi^{\alpha}\ \eta^{\beta}_{; \gamma}) u^{\gamma}[/itex]
The last step doesn't have anything to do with differentiation; it's just a fact about tensors: If [itex]w_{\alpha \beta ; \gamma}[/itex] is anti-symmetric in the first two indices, then [itex]w_{\alpha \beta ; \gamma} = 3 w_{[\alpha \beta ; \gamma]} - w_{\gamma \alpha ; \beta} - w_{\beta \gamma ; \alpha}[/itex]