As in the comments, because you are initially stationary relative to the Earth's surface, your initial velocity is exactly the same as that of the ground. The reason why is friction and air resistance: if you weren't so (perhaps you'd just dropped in from space, maybe from Betelgeuse Seven to warn Arthur Dent of a disaster in the offing, and you hadn't calculated your Earth spin velocity quite right), you'd be dragging along the ground and you'd also feel a strong headwind against your velocity (as you move relative to the atomosphere). These forces would only cease when you were moving along with the ground, i.e. stationary relative to the ground. So when you jump off the ground, your horizontal velocity component does not change, and so you track alongside the point you pushed off. It's much like two spaceships in orbit initially flying side by side with a spring between them. Imagine that one pushed off the other, and the spring thereafter brought them back together. Their horizontal velocity components don't change over the course of the thought experiment.
As for the tiny jump relative to the big Earth bit, I think you mean something like the following. On the ground, our freefalling acceleration towards the Earth centre owing to gravity alone is $g$, and the ground resists this by thrusting upwards on us. This is an accurate picture for a stationary Earth. Now for a spinning Earth: to accelerate us on a circular path so that we stay stationary relative to the ground calls for an acceleration equal to $\omega^2 \, r$, where $\omega$ is the angular speed of the Earth and $r = R_\oplus cos\theta$ our orthogonal distance to the axis of Earth rotation, $R_\oplus$ the Earth's radius and $\theta$ our latitude. Therefore this acceleration at the equator is:
$$\omega^2\,R_\oplus = \left(\frac{2\pi\text{ rad}}{24\times 3600s}\right)^2 \times 6.4\times 10^6m$$
or about $3.5\times 10^{-3}\,g$. Thus, this reduces the force that the ground must push us up with by about 0.35% - it is nothing like what is needed to reduce this force to nought. Another possible meaning is that our escape velocity i.e. the upwards velocity that would make our kinetic energy equal to the shift in gravitational potential energy $G\,M_\oplus\,m/R_\oplus$ (where $m$ is our mass) needed to reach a point infinitely far from Earth is about 11 kilometres a second. Pretty impressive compared with any plausible "jump" velocity.