The rest of the article leads me to believe that it's just like a normal derivative, except that it's being taken with respect to a function instead of a variable.
I should have mentioned this in my original post, but part of my confusion is also coming from a book I have (Classical Dynamics by Thornton and Marion), which says:
[itex]\frac{\partial \mathcal{S}}{\partial \alpha} d \alpha \equiv \delta \mathcal{S}[/itex]
where
[itex]\mathcal{S}(\alpha) = \int_{t_1}^{t_2} \! L(\mathbf{q}(\alpha, t), \dot{\mathbf{q}}(\alpha, t); t) \, dt[/itex]
and
[itex]\mathbf{q}(\alpha, t) = \mathbf{q}(0, t) + \alpha \eta(t)[/itex]
and
[itex]\eta(t)[/itex] is a function such that [itex]\eta(t_1) = \eta(t_2) = 0[/itex]
I'm having trouble seeing how these two definitions of the [itex]\delta[/itex] operator are equivalent. Thornton and Marion's [itex]\eta(t)[/itex] seems to be the same as Wikipedia's [itex]\boldsymbol\varepsilon(t)[/itex].
Does the following look correct?
Let f be a function [itex]f(x(t))[/itex].
Approaching from the Wikipedia side:
Let [itex]\varepsilon(t)[/itex] be a small perturbation from [itex]x(t)[/itex].
[itex]\delta f = f(x(t) + \varepsilon(t)) = \varepsilon(t) \frac{\partial f}{\partial x(t)}[/itex]
Approaching from the Thornton and Marion side:
[itex]x(t) = x(0, t); x(\alpha, t) = x(0, t) + \alpha \eta(t)[/itex]
[itex]\delta f = \frac{\partial f}{\partial \alpha} d \alpha = \frac{\partial f}{\partial x(t)} \frac{\partial x(t)}{\partial \alpha} d \alpha = \eta(t) \frac{\partial f}{\partial x(t)} d \alpha[/itex]
I'm not sure what to do with that [itex]d \alpha[/itex] in the Thornton and Marion approach. Should that be there?