I'm struggling a bit with Sweedler notation. Let $(H,∆,ε,S,m,u)$ be a Hopf algebra over a commutative ring $k$ and let $X,Y$ be right $H$-comodules which are finitely generated projective as $k$-modules (if you like, let $k$ be a field and let $X,Y$ be finite-dimensional vector spaces).
Then $X^*$ is also an $H$-comodule via
$$π_{X^*}(f)(x) = (f_{(1)}⊗f_{(2)})(x) = (f⊗S)∘π_X(x) = f(x_{(1)})⊗S(x_{(2)})$$
and $X⊗Y$ is an $H$-comodule via
$$π_{X⊗Y}(x⊗y) = x_{(1)}⊗y_{(1)}⊗m(x_{(2)}⊗y_{(2)}).$$
The source I am using says that the coaction on $X^*⊗X$ is given by
$$π_{X^*⊗X}(f⊗x) = f⊗x_{(1)}⊗m(S(x_{(2)})⊗x_{(3)}) $$
which I think means that symbols $x_{(2)},x_{(3)}$ refer to (sets of) elements of $H$ that "split off" from $f$ and $x$ respectively under the coactions, but this is confusing to me because there also seems to be an implicit identification $X^*⊗H = \mathrm{Hom}_k(X,H)$ floating around. And then to show that the evaluation map is an $H$-comodule homomorphism we have
$$ (\mathrm{ev}_X⊗\mathrm{id}_H)∘π_{X^*⊗X}(f⊗x) = f(x_{(1)})⊗S(x_{(2)})x_{(3)} \\ = f(x_{(1)})⊗ε(x_{(2)}) = f\big(x_{(1)}ε(x_{(2)})\big)⊗1_H = f(x). $$
Why should the components $x_{(1)},x_{(2)},x_{(3)}$ reassemble like this to produce $x$ when it looks like they didn't all come from $x$ to begin with? In other words, this is completely fine if $x_{(1)}⊗x_{(2)}⊗x_{(3)} = (π_X⊗\mathrm{id}_H)∘π_X(x)$, but I don't see why that should be the case.
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.