When heat is increased, does water move upward in a straw due to increased capillary action or increased pressure?
It almost certainly was due to thermal expansion of water and the air above it. Consider that the thermal expansion coefficient of water near room temperature is ~$0.000214$ K$^{-1}$. [Though it is not linear over a broad range, assume for the small temperature rise in your car that it is.]

If the water was cold (e.g. ~$277$ K, maximum density) and it warmed to ~$300$ K ($80$o F), a temperature change of ~$23$ K, it would have expanded by a factor $(300 \ce{K} - 277 \ce{K}) \cdot 0.000214 \ce{K^{-1}} = 1.005$.

Assume you have a one-liter bottle filled completely with water when you start; by the time you return it's $1.005$ L. In the whole bottle, the change would hardly be noticeable, but the narrow straw greatly magnifies the rise of liquid. If the straw is $6$ mm in diameter, the volume of liquid is given as: $$V=\pi\cdot r^2\cdot h$$ 

$$5 \mathrm{mL} = 5 \ce{cm}^3$$

$$r = 0.3 \ce{cm}$$

$$h = \frac{5 \ce{cm}^3}{\pi \cdot (0.3 \ce{cm})^2} = 18 \ce{cm}$$

So the water would have risen about $18$ cm in the straw!

This exercise only took into account the water's expansion... for any air above the water in the bottle, and for other (approximately) ideal gases, $PV=nRT$ and thermal expansion would be far greater.

To some extent, the bottle itself expands, decreasing the rise of the liquid, so these calculations are not very accurate. Most likely, air expansion contributed more to the "mass ejection" than did that of water... it depends on whether you view your container as half-full or half-empty.