The general underlying fact is the component is a conductive element continuing the transmission lines of the connecting traces.
So, whatever the width and height of the conductor over the substrate is, that's your transmission line impedance, and component length is line length.
So, bigger components have a proportionally lower cutoff frequency, and wider components have a lower Zo.
What does Zo do? The ratio between rated value and Zo is relevant. The closer they match, the nearer to 1 the ratio is -- the higher the cutoff frequency will be. The more mismatched they are, the lower the cutoff will be -- and approximately proportionally so.
So a 100 ohm resistor will be a good match to most traces, and itself over the substrate, while a 10 or 1k resistor will have about a 10x lower cutoff.
Which kind of cutoff will it have? R < Zo, inductive; R > Zo, capacitive. That is, for lower impedances, the equivalent inductance dominates, while for higher impedances, the equivalent capacitance dominates.
We don't need to do any messy transmission line calculations if we're only after the low-frequency equivalent:
ESL = mu_0 * Z/Zo * length
Cp = e_0 * k * Zo/Z * length
(assuming substrates with mu_r = 1, and some dielectric constant k, the usual case)
Where mu_0 ~= 1.257 uH/m, e_0 ~= 8.84 pF/m, Z is the transmission line impedance and Zo is the impedance of free space, ~377 ohms (also Zo = sqrt(mu_0 / e_0) if you prefer reducing it in terms of just the two constants, but this is easier to remember I'd say).
Including the crazy high frequency range documented in the Vishay document, you'll have to consider full transmission line behavior, as well as width, height, material (the resistor body is usually Al2O3, while the substrate is whatever it is), higher order modes... radiating modes probably?, whatever. That's unsurprising, there's just so many more degrees of freedom for high frequency waves so we aren't going to get away so easily with a hand-wave up there.
Tim