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Calling B's corresponding duration $\frac{2 \, BP}{c} \frac{1}{1 - \beta^2} = \mathop{\Delta \, \tau_B}\limits_{\text{ping trip } A}$ therefore

$\mathop{\Delta}\limits_{\text{ping}} \tau_A[ W ] = \mathop{\Delta \, \tau_B}\limits_{\text{ping trip } A} \sqrt{ 1 - \beta^2 }$, as may have been expected.

Well, shouldn't that have been pretty much the derivation I just sketched?.

Calling B's corresponding duration $\frac{2 \, BP}{c} \frac{1}{1 - \beta^2} = \mathop{\Delta \, \tau_B}\limits_{\text{ping trip } A_W}$ therefore

$\mathop{\Delta}\limits_{\text{ping}} \tau_A[ W ] = \mathop{\Delta \, \tau_B}\limits_{\text{ping trip } A_W} \sqrt{ 1 - \beta^2 }$, as may have been expected.

Well, shouldn't that have been pretty much the derivation I just sketched?.

Edit

For completeness, and to emphasize a particular point in the following, here's also the derivitation involving light clocks "perpendicular to the direction of motion" (which seems to have been mentioned in passing in the OP's question):

Expanding on the setup described above, with the principal protagonists A and B and suitable auxiliary participants (W and J at rest wrt. A; P and Q at rest wrt. B), and all of them "sitting or moving in one line", we now also consider

• participant F (at rest wrt. A, J, W) with distance ratios $\left( \frac{AF}{FJ} \right)^2 + \left( \frac{AJ}{FJ} \right)^2 = 1$,
  and (without loss of generality, but just to re-use setup relations from above) with $\frac{AW}{FJ} = 1$,
  therefore $\frac{AF}{FJ} = \sqrt{ 1 - \left( \frac{AJ}{FJ} \right)^2 } = \sqrt{ 1 - \left( \frac{AJ}{AW} \right)^2 } = \sqrt{ 1 - \beta^2 }$; and

• participant G (at rest wrt. B, P, Q) with distance ratios $\left( \frac{BG}{GP} \right)^2 + \left( \frac{BP}{GP} \right)^2 = 1$,
  and such that G and F met each other in passing.

Importantly, the entire region containing the setup is of course supposed to be flat. Therefore it can be demonstrated (what otherwise may be glanced over for seeming "too obvious to even point out"), that

• F's indication of having been passed by G was simultaneous to A's indication of having been passed by B; and vice versa that

• G's indication of having been passed by F was simultaneous to B's indication of having been passed by A.

Then, by the same argument that was used above for comparison of distance ratios between pairs of participants who were not at rest to each other, we set:

$\frac{AF}{BG} = \frac{BG}{AF}$, and therefore $\frac{AF}{BG} = 1.$

With

$\mathop{\Delta , \tau_A}\limits_{\text{ping trip } B_G} = \frac{2 \, FJ}{c} = \frac{2 \, AF}{c} / \sqrt{ 1 - \beta^2 }$ and

$\mathop{\Delta}\limits_{\text{ping}} \tau_B[ G ] = \frac{2 \, BG}{c}$ follows

$\mathop{\Delta}\limits_{\text{ping}} \tau_B[ G ] = \mathop{\Delta \, \tau_A}\limits_{\text{ping trip } B_G} \sqrt{ 1 - \beta^2 }$.

Finally, as can be shown explicitly, it holds symmetrically that

$\mathop{\Delta}\limits_{\text{ping}} \tau_A[ F ] = \mathop{\Delta \, \tau_B}\limits_{\text{ping trip } A_F} \sqrt{ 1 - \beta^2 }$.

• B and A were passing each other (which is supposed to be visible by everyone else; a.k.a. "emitting a light signal"),

• there exists some participant (let's call it Q) who was and remained at rest wrt. B, who therefore of course was passed by W "sometime", and specificly: whose indication of being passed by W was simultaneous to B's indication of being passed by A, and

• there exists some participant (let's call it Q) who was and remained at rest wrt. B (as well as P) and who was passed W just as Q and W observed (together, at their meeting) that B and A had passed each other.

• B and A were passing each other (which is supposed to be visible by everyone else; a.k.a. "emitting a light signal"),

• there exists some participant (let's call it P) who was and remained at rest wrt. B, who therefore of course was passed by W "sometime", and specificly: whose indication of being passed by W was simultaneous to B's indication of being passed by A, and

• there exists some participant (let's call it Q) who was and remained at rest wrt. B (as well as P) and who was passed W just as Q and W observed (together, at their meeting) that B and A had passed each other.

... a.k.a. "ping duration $\mathop{\Delta}\limits_{\text{ping}} \tau_A[ W ] = \frac{2 \, AW}{c}$.

... a.k.a. "ping duration" $\mathop{\Delta}\limits_{\text{ping}} \tau_A[ W ] = \frac{2 \, AW}{c}$.