Long story short: it seems there is a mistake in your source for $\pi_{X^* \otimes X}$, and also the employed Sweedler notation was quite obfuscating.
Let me use the (very common) Sweedler notation $\pi_X(x) = x_{[0]} \otimes x_{[1]}$ for the right coactions.
Here the 0-indexed factor is always in the comodule, while the ones indexed with 1, 2, ... are in $H$.
For iterated coactions, I write $x_{[0,0]} \otimes x_{[0,1]} \otimes x_{[1]} = x_{[0][0]} \otimes x_{[0][1]} \otimes x_{[1]}$, and with this we can express coassociativity as $x_{[0,0]} \otimes x_{[0,1]} \otimes x_{[1]}
= x_{[0} \otimes x_{[1](1)} \otimes x_{[1](2)}$, where the indicies in parentheses are the usual Sweedler notation for the Hopf algebra $H$.
As you said, on tensor products, the coaction is defined as
$$
\pi_{X \otimes Y}(x \otimes y) = x_{[0]} \otimes y_{[0]} \otimes x_{[1]} y_{[1]}.
$$
In terms of maps,
$$
\pi_{X \otimes Y} = X \otimes Y \otimes m \circ X \otimes \tau_{H, Y} \otimes H
\circ \pi_X \otimes \pi_Y,
$$
where $\tau$ is the tensor factor swap in the category of vector spaces, i.e. $\tau_{X, Y}(x \otimes y) = y \otimes x$.
And for the dual space, $\pi_{X^*}(f) \in X^* \otimes H \cong Hom_k(X, H)$ via
$$
\pi_{X^*}(f)(x) = f(x_{[0]}) S(x_{[1]}),^{1}
$$
so that $\pi_{X^*}(f) = f \otimes S \circ \pi_X$.
Putting this together, we have
\begin{align*}
\pi_{X^* \otimes Y}(f \otimes y)
&= X \otimes Y \otimes m \circ X \otimes \tau_{H, Y}
\circ (f \otimes S \circ \pi_X) \otimes \pi_Y(y).
\end{align*}
Thus, identifying $X^* \otimes Y \otimes H \cong Hom_k(X, Y \otimes H)$, we have
$$
\pi_{X^* \otimes Y}(f \otimes y)(x) = f(x_{[0]}) y_{[0]} \otimes S(x_{[1]}) y_{[1]}
$$
And then we find for example
\begin{align*}
(\operatorname{ev}_X \otimes H \circ \pi_{X^* \otimes X})(f \otimes x)
&= \operatorname{ev}_X \otimes H \circ
X^* \otimes X \otimes m \circ X^* \otimes \tau_{H, X}
\circ (f \otimes S \circ \pi_X) \otimes \pi_X(x) \\
&= m \circ \operatorname{ev}_X \otimes H \otimes H
\circ X^* \otimes \tau_{H, X} \otimes H
\circ (f \otimes S \circ \pi_X) \otimes \pi_X(x) \\
&= f(x_{[0,0]}) S(x_{[0,1]}) x_{[1]} \\
&\overset{*}{=} f(x_{[0]}) S(x_{[1](1)}) x_{[1](2)} \\
&= f(x_{[0]}) \varepsilon(x_{[1]}) 1_H \\
&= f(x) 1_H \\
&= \pi_{k} \circ \operatorname{ev}_X(f \otimes x)
,
\end{align*}
so that evaluation indeed is a comodule map
(the mysterious-seeming third equality is best seen using string diagrams, really${}^2$).
The step marked $*$ uses coassociativity of the coaction as described above.
${}^1$ For completeness:
In terms of maps, we have
$$
\pi_{X^*} = \tau_{H, X^*} \circ S \otimes X^*
\circ \operatorname{ev}_X \otimes H \otimes X^* \circ X^* \otimes \pi_X \otimes X^* \circ X^* \otimes \operatorname{coev}_X.
$$
EDIT
${}^2$ To be more transparent, I'll prove it algebraically using the zig-zags of evaluation and coevalutation.
I sometimes suppress the tensor product symbol.
\begin{align*}
&\operatorname{ev}_X \otimes H \circ \pi_{X^* \otimes X} \\
&=
\operatorname{ev}_X \otimes H \circ
X^* X \otimes m \circ X^* \otimes \tau_{H, X} \otimes H
\circ \pi_{X^*} \otimes \pi_X \\
&=
m \circ \operatorname{ev}_X \otimes H H \circ X^* \otimes \tau_{H, X} \otimes H
\circ \pi_{X^*} \otimes \pi_X
\\
&=
m
\circ \operatorname{ev}_X \otimes H H
\circ X^* \otimes \tau_{H, X} \otimes H
\circ \tau_{H, X^*} \otimes X H \\ & \quad
\circ S \otimes X^* X H
\circ \operatorname{ev}_X \otimes H X^* X H
\circ X^* \otimes \pi_X \otimes X^* X H \\ & \quad
\circ X^* \otimes \operatorname{coev}_X \otimes X H
\circ X^* \otimes \pi_X
\end{align*}
Up until here, I have just inserted things, and pulled the evaluation past the
multiplication.
Now note that
$$
\operatorname{ev}_X \otimes H H \circ X^* \otimes \tau_{H, X} \otimes H \circ \tau_{H, X^*} \otimes X H = H \otimes \operatorname{ev}_X \otimes H.
$$
Then we get the following, to which we apply the zigzag (underlined, and after pulling
the left-most evaluation past a lot of things)
\begin{align*}
&=
m
\circ H \otimes \operatorname{ev}_X \otimes H \\ & \quad
\circ S \otimes X^* X H
\circ \operatorname{ev}_X \otimes H X^* X H
\circ X^* \otimes \pi_X \otimes X^* X H \\ & \quad
\circ X^* \otimes \operatorname{coev}_X \otimes X H
\circ X^* \otimes \pi_X
\\
&=
m
\circ S \otimes H
\circ \operatorname{ev}_X \otimes H H
\circ X^* \otimes \pi_X \otimes H
\\ & \quad
\circ \underline{X^* X \otimes \operatorname{ev}_X \otimes H
\circ X^* \otimes \operatorname{coev}_X \otimes X H}
\circ X^* \otimes \pi_X
\\
&=
m
\circ S \otimes H
\circ \operatorname{ev}_X \otimes H H
\circ X^* \otimes \pi_X \otimes H
\circ X^* \otimes \pi_X
\\
&=
m
\circ S \otimes H
\circ \operatorname{ev}_X \otimes H H
\circ X^* \otimes \pi_X \otimes H
\circ X^* \otimes \pi_X
.
\end{align*}
This maps
\begin{align*}
f \otimes x
&\mapsto f \otimes x_{[0]} \otimes x_{[1]} \\
& \mapsto f \otimes x_{[0,0]} \otimes x_{[0,1]} \otimes x_{[1]} \\
& \mapsto f (x_{[0,0]}) x_{[0,1]} \otimes x_{[1]} \\
& \mapsto f (x_{[0,0]}) S(x_{[0,1]}) x_{[1]},
\end{align*}
as claimed above.
