After considering @benrg's comments, I realize that my first answer contained too strong statements about the relation between the two redshifts. I try here to moderate my answer, but you might want to accept their answer instead.
It is common to think of the two redshifts as having nothing to do with each other. Doppler shifts arise when the observer and/or the emitter moves through space, whereas the cosmological redshift can be derived considering stationary emitters and stationary observers in an expanding space.
Because the cosmological redshift doesn't involve movement through space, it is often considered completely different from the Doppler. However, it is also possible to derive the cosmological redshift by considering it as infinitely many infinitesimally small Doppler shift (e.g. Lewis 2016). I admit that I'm not well enough versed in general relativity to be certain about my statements, but just because an infinitesimally small patch of spacetime is flat doesn't necessarily mean that infinitely many such patches add up to be flat. However, as @benrg says, in GR there is only one redshift.
Different or the same?
The reason I think it makes sense to view the Doppler shift and the cosmological redshift as two separate mechanisms is the following:
In principle you could have a universe (non-capitalized, since it's not our universe, the Universe) that were static when a distant galaxy emitted a photon, then at some point expanded quickly by a factor of 2, and then again is static. In this hypothetical case, the observer would still measure the photon to have been redshifted by a factor of 2 (i.e. $z=1$).
That this is true can bee seen from considering the mathematical derivation of the cosmological redshift (see e.g. here) which involves an integral, the result of which only depends on the initial and the final state, not on the expansion history.
In contrast, if you and your friend stand still with respect to each other while your friend shines her flashlight at you, then run away from each other with a relative velocity of $0.6c$, then stand still before you receive the light (i.e. analogously to the hypothetical universe above), then you would measure no redshift; you wouldn't measure the special relativistic Doppler shift of $z+1 = \sqrt{\frac{1+v/c}{1-v/c}} = 2$ that you would if you were receding from each other while either emitting or observing.
In the real Universe, galaxies move through space (i.e. they change their comoving coordinate $\chi$), and space expands (i.e. the scale factor $a$ evolves). If the physical distance to a galaxy is
$$
d = a \chi,
$$
then the change in this distance gives their total velocity wrt. us, and is obtained through differentiation:
$$
\begin{array}{rcl}
v_\mathrm{tot} & = & \dot{a}\chi + a\dot{\chi} \\
& \equiv & v_\mathrm{rec} + v_\mathrm{pec},
\end{array}
$$
where dots denote differentiation with respect to time, and the two terms have been identified as the cosmological recession velocity, and the peculiar, "normal" velocity. Each of these terms give rise to a redshift, but through two very different mechanisms. Only the latter term is called a Doppler shift.
