In the symmetric case where both experience the same turn-around force they both end up the same age and (each having seen the other age by the same amount overall) are not surprised by that result. Here what they actually see is the other ageing more slowly on the outward trip and more quickly on the return (due to the Doppler effect). But what they infer after correcting for light travel times is that the other was ageing more slowly on both legs of the trip but more rapidly during the turn-around.
But in the usual asymmetric formulation where only one is accelerated, the Doppler effects during separation and return are not quite the same – with the homie seeing fewer ticks on the traveller’s clock than vice versa (I’m just telling you this rather than showing how it actually works out, but if you do work through it you’ll find that it does). And if they both do the signal-travel-time corrections, then what the homie will infer is that the traveller was ticking slowly throughout while the traveller will infer that the homie was ticking slower during the constant speed legs of the trip but speeded up during the turn around by twice the amount needed to cancel the slow periods. So either way they both end up agreeing that the traveller’s clock has ticked fewer times.