The function $q$ strictly decreases along the solutions of the gradient flow outside of the zero section. Hence any orbit that starts outside the zero section will not converge to a critical point in backwards time, and does not show up in some moduli space of orbits connecting critical points. The orbits on the zero section correspond bijectively to orbits on $M$.

If one chooses orientations of all the unstable manifolds in $M$, this naturally also gives you orientations for all the unstable manifolds of the critical points in $E$, hence even the orientations of the moduli spaces on $M$ and $E$ agree.

Much more interesting is to not take the function you give, but the function

$$ f^2_E(v_m)=f(m)-q(v_m) $$

The critical points on $E$ correspond to the ones on $M$ shifted by the dimension of $E$, as there are this many more unstable directions. The moduli spaces on $M$ and $E$ again correspond bijectively to each other, but the orientations might not agree. However, if $E$ is an oriented vector bundle, we can orient the unstable manifolds of the critical points on $E$ taking into account this orientation on the fibers. Doing this correctly will show that the moduli spaces are isomorphic as oriented manifolds. This will give an isomorphism

$$ HM_*(M)\cong HM_{*+\dim E}(E,f_E^2) $$

which is the Morse theoretic Thom isomorphism. This can be found for example in the appendix of this paper http://arxiv.org/abs/0810.1995 by Abbondandolo and Schwarz

To see that the orientation assumptions are necessary, you can compute this explicitly for the Mobius strip.

The function $q$ strictly decreases along the solutions of the gradient flow outside of the zero section. Hence any orbit that starts outside the zero section will not converge to a critical point in backwards time, and does not show up in some moduli space of orbits connecting critical points. The orbits on the zero section correspond bijectively to orbits on $M$ if you choose a metric which respects a horizontal-vertical splitting of $TE$ in the obvious way.

If one chooses orientations of all the unstable manifolds in $M$, this naturally also gives you orientations for all the unstable manifolds of the critical points in $E$, hence even the orientations of the moduli spaces on $M$ and $E$ agree.

Much more interesting is to not take the function you give, but the function

$$ f^2_E(v_m)=f(m)-q(v_m) $$

The critical points on $E$ correspond to the ones on $M$ shifted by the dimension of $E$, as there are this many more unstable directions. The moduli spaces on $M$ and $E$ again correspond bijectively to each other, but the orientations might not agree. However, if $E$ is an oriented vector bundle, we can orient the unstable manifolds of the critical points on $E$ taking into account this orientation on the fibers. Doing this correctly will show that the moduli spaces are isomorphic as oriented manifolds. This will give an isomorphism

$$ HM_*(M)\cong HM_{*+\dim E}(E,f_E^2) $$

which is the Morse theoretic Thom isomorphism. This can be found for example in the appendix of this paper http://arxiv.org/abs/0810.1995 by Abbondandolo and Schwarz

To see that the orientation assumptions are necessary, you can compute this explicitly for the Mobius strip.