Let $H$ be a Hopf algebra, and $V$ and $W$ a left, and a right,
$H$comodule respectively. The tensor product
$$
V \otimes W
$$
has an obvious $H$$H$bicomodule structure.
If $V$ and $W$ are irreducible as left and right comodules, then
is $V \otimes W$ irreducible as a $H$$H$bicomodule?
No, off course, not. You need a slightly stronger condition. One of your comodules need to be absolutely irreducible.
To prove it, apply the fundamental theorem of coalgebra. The comodules give you simple subcoalgebras $C_M$ and $C_N$ in the coradical of $H$. You are asking for the algebra $(C^{\ast}_M)^{op}\otimes C^{\ast}_N$ to be simple.
For an elementary example let ${\mathbb R}$ be the ground field, $H$ the free cocommutative Hop algebra generated by the trigonometric coalgebra ${\mathbb C}^{\ast}$. Let $M=N={\mathbb C}$ be the $\mathbb C$module, hence, ${\mathbb C}^\ast$comodule, hence $H$comodule. The $H$bicomodule structure on $M\otimes N$ is essentially its module structure over ${\mathbb C}\otimes_{\mathbb R}{\mathbb C}\cong {\mathbb C}\oplus{\mathbb C}$. This is not irreducible because $dim(M\otimes N)=4$ but its simple modules are 2dimensional.

$\begingroup$ Sorry but I don't see why what I am asking for is equivalent to $(C^*_M)^{op} \otimes C^*_N$ being simple. (Also, I guess $C^*_M$ denotes the linear dual of $C^*_M$ endowed with convolution multiplication?) $\endgroup$ – Jake Wetlock Apr 14 '20 at 11:29

$\begingroup$ You are asking for $M\otimes N$ to be a simple $(C^{\ast}_M)^{op}\otimes C^{\ast}_N$module. And you know that $M\otimes N$ is a faithful finitedimensional $(C^{\ast}_M)^{op}\otimes C^{\ast}_N$module. Now simplicity of the algebra $(C^{\ast}_M)^{op}\otimes C^{\ast}_N$ implies simplicity of the module $M\otimes N$. In the opposite direction, hit it with Jacobson Density Theorem. $\endgroup$ – Bugs Bunny Apr 14 '20 at 11:59

$\begingroup$ Yes or no, since you have a typo: $C^{\ast}_M$ is the linear dual of coalgebra $C_M$. $\endgroup$ – Bugs Bunny Apr 14 '20 at 12:01

$\begingroup$ Yes, it should read "$C^*_M$ denotes the linear dual of $C_M$". $\endgroup$ – Jake Wetlock Apr 14 '20 at 13:04