Addendum to the above:

Given a \\(\mathcal{V}\\)-functor \\(F : \mathcal{Z} \to \mathcal{Y}\\), define \\(F^\text{op} : \mathcal{Z}^\text{op} \to \mathcal{Y}^\text{op}\\) by \\(F^\text{op}(z) = F(z)\\). It's easy to prove \\(F^\text{op}\\) is indeed a \\(\mathcal{V}\\)-functor.

Note that \\(\check{F}(y, z) = \mathcal{Y}(y, Fz) = \mathcal{Y}^\text{op}(F^\text{op} z, y) = \widehat{F^\text{op}}(z, y)\\).

So we can use this construction to turn conjoints into companions and vice versa. In particular we have

\[\check{GF}(x, z) = \widehat{(GF)^\text{op}}(z, x) = \widehat{G^\text{op} F^\text{op}}(z, x) = (\widehat{G^\text{op}}\circ \widehat{F^\text{op}})(z, x) = (\check{F}\circ\check{G})(x, z)\]

... which gives us the rule about composing conjoints directly from the one about composing companions.

[EDIT to add: this turned out to be a little more complex than I'd initially thought, I think it's right now but the last step needs proving]

Given a \\(\mathcal{V}\\)-functor \\(F : \mathcal{Z} \to \mathcal{Y}\\), define \\(F^\text{op} : \mathcal{Z}^\text{op} \to \mathcal{Y}^\text{op}\\) by \\(F^\text{op}(z) = F(z)\\). It's easy to prove \\(F^\text{op}\\) is indeed a \\(\mathcal{V}\\)-functor.

Note that \\(\check{F}(y, z) = \mathcal{Y}(y, Fz) = \mathcal{Y}^\text{op}(F^\text{op} z, y) = \widehat{F^\text{op}}(z, y)\\).

So we can use this construction to turn conjoints into companions and vice versa. In particular we have

\[\check{GF}(x, z) = \widehat{(GF)^\text{op}}(z, x) = \widehat{G^\text{op} F^\text{op}}(z, x) = (\widehat{G^\text{op}}\circ \widehat{F^\text{op}})(z, x) = (\check{F}\circ\check{G})(x, z)\]

... which gives us the rule about composing conjoints directly from the one about composing companions.

[EDIT to add: this turned out to be a little more complex than I'd initially thought, I think it's right now but the last step needs proving]