If $N$ is a subcomodule of $M$, then the H-submodule generated by $N$ is a H-Hopf-module of $M$. Let $H$ be a Hopf algebra with $S$ its antipode and let $M$ be a right $H$-Hopf-module. We will write $\rhd$ the action of the left $H^*$-module on $M$ induced by the structure of right $H$-module, that is, $h^*\rhd m=\Sigma_{(m)}\langle h^*,m_{(1)} \rangle m_{(0)}$, $\forall m\in M, h^*\in H$. Let $\rightharpoonup$ the action of $H$ on $H^*$ such that $\langle a \rightharpoonup f, b \rangle =\langle f,ba \rangle$. I know that:
$$
h^*\rhd(m\cdot h)=\sum_{(h)}((h_{(2)}\rightharpoonup h^*)\rhd m)\cdot h_{(1)}
$$

I want to show that if $N$ is a subcomodule of $M$, then the H-submodule generated by $N$ is a H-Hopf module of $M$. 

The only book that I found this said that the information is clear and follows by the formula above (that I proved in an exercise), but I don't see why it is clear.
 A: Note that it suffices to show that the $H$-submodule $N'$ generated by $N$ is a $H$-subcomodule ($H^*$-submodule), as it is already a submodule and the compatibility condition is valid on all of $M$.
I think there is something wrong about the compatibility you derive. The correct compatibility of the right $H$-action and coaction should be
$$
\delta(m\cdot h)=\sum_{m\cdot h}(n\cdot h)_{(0)}\otimes (m\cdot h)_{(1)}=\sum_{(h),(m)}m_{(0)}\cdot h_{(1)}\otimes m_{(1)}h_{(2)}
$$
To explain the notation, $\delta\colon M\to M\otimes H$ is the right $H$-coaction and can be written using Sweedler's notation as $$\delta(m)=\sum_{(m)}m_{(0)}\otimes m_{(1)}.$$
Now we use the dual coaction of $H^*$ to translate this into
$$
h^*\rhd(m\cdot h)=\sum_{(h^*)}({h^*}_{(2)}\rhd m)\cdot({h^*}_{(1)}\rightharpoonup h).
$$
Now, $N'$ is generated by elements of the form $n\cdot h$ for $h\in H$, $n\in N$. The compatibility condition of Hopf modules gives an expression for the coproduct of such elements: 
$$
\delta(n\cdot h)=\sum_{n\cdot h}(n\cdot h)_{(0)}\otimes (n\cdot h)_{(1)}=\sum_{(h),(n)}n_{(0)}\cdot h_{(1)}\otimes n_{(1)}h_{(2)}\in N'\otimes H
$$
Note that this proves that $N'$ is also a $H$-subcomodule, i.e. $\delta(N')\leq N'\otimes H$. On one side we have the coaction applied to $n\cdot h$ and on the other side we have a sum of tensors in $N'\otimes H$.
Similarly, this can be see using the dual action: On the LHS of the compatibility we act on a generating element of $N'$, and the RHS is in $N'$ as $N$ is a $H$-submodule.
