My favorite approach is the Bondi k-calculus (which is of course just algebra, not calculus). See for instance
https://en.wikipedia.org/wiki/Bondi_k-calculus
The doppler shift factor is just "k". We'll work an example where k=10 for simplicity.
This corresponds to a velocity of (k^2 - 1) / (k^2 + 1) * c, or 99/101 of the speed of light, approximately .98 c.
If an object travels with this velocity for 100 seconds in its own frame away from a beacon that emits one pulse a second, for 100 seconds (again, as measured by its own clock) it will receive 10 pulses on the outbound trip. This happens because the beacon is redshifted so it appears to emit one pulse every 10 seconds, and 1 pulse every 10 seconds for 100 seconds is 10 pulses.
[add]For some reason , when I wrote this, I called the "beacon" frame the "earth" frame, and the object the "spaceship" frame. SOrry for being a bit careless with the analogy.
On the inbound trip, the doppler shift will multiply the frequency by 10, rather than divide it by 10, so the ship will receive 1000 pulses on the inbound trip.
Thee total number of pulses received, on the round trip is the sum of the number of pulses received on the inbound trip plus the number of pulses received on the outbound trip, for a total of 1010 pulses, received over 200 seconds.
This is what happens in the object's frame. Now, what happens in the Earth frame?
The time dilation factor is 1/sqrt(1-v^2/c^2), this turns out to be alternatively and more easily expressed as being equal to (k + 1/k)/2, which turns out to be 5.05. So, in the Earth frame the ship travels outward for 505 seconds, not 100 seconds. And it travels inbound for 505 seconds. So the total travel time is 1010 seconds, during which 1010 pulses are emitted.
So, everyone agrees that the beacon emits 1010 pulses. In the beacon frame, this occurs over 1010 seconds, in the object frame it occurs in 200 seconds, due to the time dilation factor of slightly over 5:1.