= Solution
A useful <comodule natural transformation formula> determines $\beta$ from a single <linear functional>. Define
$$
\alpha=\varepsilon_H\beta_H:H\to k,
$$
using the regular right <comodule> $H$. For any <vector space> $V$, the cofree right <comodule> $V\otimes H$ has coaction $1_V\otimes\Delta$. <Naturality> with respect to all maps $h\mapsto v\otimes h$ gives
$$
\beta_{V\otimes H}=1_V\otimes\beta_H.
$$
The coaction $\rho_M:M\to M\otimes H$ is itself a morphism of right <comodules>. Its <naturality> equation, followed by $1_M\otimes\varepsilon_H$, therefore gives
$$
\boxed{\beta_M(u)=\sum u_{(0)}\alpha(u_{(1)}).}
$$
This derivation works for all <comodules>, not merely finite-dimensional ones.
The monoidal equation $\beta_{M\otimes N}=\beta_M\otimes\beta_N$, evaluated on the two regular <comodules> and followed by their <counits>, implies
$$
\alpha(ab)=\alpha(a)\alpha(b).
$$
The unit equation gives $\alpha(1)=1$. Thus $\alpha$ is a unital <algebra homomorphism over a field> $H\to k$. In addition, the $K$-colinearity of $\beta_H$ gives the useful intertwining relation
$$
\sum g(h_{(1)})\alpha(h_{(2)})=\sum\alpha(h_{(1)})f(h_{(2)}).
$$
This fixes the orientation of $\alpha$ relative to $f,g$.
For the <convolution product for coalgebra maps>, define $\alpha^{-1}=\alpha S$. Multiplicativity of $\alpha$ and the two <antipode> identities show
$$
\sum\alpha(h_{(1)})\alpha(S(h_{(2)}))=\varepsilon(h),\qquad\sum\alpha(S(h_{(1)}))\alpha(h_{(2)})=\varepsilon(h).
$$
Consequently
$$
\boxed{\beta_M^{-1}(u)=\sum u_{(0)}\alpha(S(u_{(1)})).}
$$
Coassociativity and the two convolution identities verify both composites directly. Since $\beta_M$ was a $K$-comodule morphism, its linear inverse is also a $K$-comodule morphism. Inverting the <naturality> and monoidal equations shows that the inverses form a monoidal <natural transformation> $g_*\Rightarrow f_*$. \b[Every such monoidal transformation is therefore invertible], without requiring a bijective antipode.
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