Alternative Cotensor Definition Let $H$ be a Hopf algebra, and $(M,\rho_r)$ and $(N,\rho_l)$ right and left $H$-comodules respectively. As usual, we define their cotensor product  to be 
$$
M \square_H N := \text{ker}\{(\rho_r \otimes \text{id} - \text{id} \otimes \rho_l ): M \otimes N \to M \otimes H \otimes N\}.
$$
I have seen an alternative version of the definition
$$
M \square_H N := \{\sum_i m_i \otimes n_i, \sum (m_i)_0 \otimes (n_i)_0 \otimes (m_i)_1 (S(n_i)_{-1}) = \sum_i m_i \otimes n_i \otimes 1 \}.
$$
Are they equivalent, and if so, then how does one show this?
 A: Yes, they are equivalent.
Let me restate this fact in more proof-friendly terms:
Let $\mathbf{k}$ be a commutative ring. All tensor signs $\otimes$ will be
understood to mean $\otimes_{\mathbf{k}}$. Let $H$ be a $\mathbf{k}$-Hopf
algebra with antipode $S:H\rightarrow H$. Let $\left(  M,\rho_{r}\right)  $ be
a right $H$-module (so that $\rho_{r}:M\rightarrow M\otimes H$ is a
$\mathbf{k}$-linear map satisfying certain axioms). Let $\left(  N,\rho_{\ell
}\right)  $ be a left $H$-module (so that $\rho_{\ell}:N\rightarrow H\otimes
N$ is a $\mathbf{k}$-linear map satisfying certain axioms). We shall use the
sum-free Sweedler notation: e.g., we write $h_{\left(  1\right)  }\otimes
h_{\left(  2\right)  }$ for the image of a $h\in H$ under the comultiplication
of $H$; we will write $m_{\left(  0\right)  }\otimes m_{\left(  1\right)  }$
for $\rho_{r}\left(  m\right)  $ when $m\in M$; and we will write $n_{\left(
-1\right)  }\otimes n_{\left(  0\right)  }$ for $\rho_{\ell}\left(  n\right)
$ when $n\in N$.
Define a $\mathbf{k}$-linear map $\Psi:M\otimes N\rightarrow M\otimes H\otimes
N$ by the requirement that
$\Psi\left(  m\otimes n\right)  =m_{\left(  0\right)  }\otimes m_{\left(
1\right)  }S\left(  n_{\left(  -1\right)  }\right)  \otimes n_{\left(
0\right)  }$ for all $m\in M$ and $n\in N$.
Define a $\mathbf{k}$-linear map $\Omega:M\otimes N\rightarrow M\otimes
H\otimes N$ by the requirement that
$\Omega\left(  m\otimes n\right)  =m\otimes1\otimes n$ for all $m\in M$ and
$n\in N$.
Let $\tau_{N}:H\otimes N\rightarrow N\otimes H$ be the twist map (i.e., the
$\mathbf{k}$-linear map sending each $h\otimes n$ to $n\otimes h$)
Your second definition of $M\square_{H}N$ then rewrites as follows:
$M\square_{H}N=\ker\left(  \left(  \operatorname*{id}\nolimits_{M}\otimes
\tau\right)  \circ\left(  \Psi-\Omega\right)  \right)  $
(I will explain more carefully below why this is equivalent to your second definition).
Since $\operatorname*{id}\nolimits_{M}\otimes\tau$ is an isomorphism of
$\mathbf{k}$-modules, this is equivalent to
(1) $M\square_{H}N=\ker\left(  \Psi-\Omega\right)  $.
So we must prove (1).
Let $\mu$ denote the multiplication map $H\otimes H\rightarrow H$ of the
$\mathbf{k}$-algebra $H$. Let $\Delta$ and $\varepsilon$ denote the
comultiplication and the counity of the $\mathbf{k}$-coalgebra $H$. We claim that
(2) $\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes
\operatorname*{id}\nolimits_{N}\right)  \circ\left(  \operatorname*{id}
\nolimits_{M\otimes H}\otimes\rho_{\ell}\right)  \circ\Psi=\rho_{r}
\otimes\operatorname*{id}\nolimits_{N}$
as $\mathbf{k}$-linear maps $M\otimes N\rightarrow M\otimes H\otimes N$.
Indeed, for any $m\in M$ and $n\in N$, we have
$\left(  \left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes
\operatorname*{id}\nolimits_{N}\right)  \circ\left(  \operatorname*{id}
\nolimits_{M\otimes H}\otimes\rho_{\ell}\right)  \circ\Psi\right)  \left(
m\otimes n\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id}
\nolimits_{N}\right)  \left(  \left(  \operatorname*{id}\nolimits_{M\otimes
H}\otimes\rho_{\ell}\right)  \left(  \underbrace{\Psi\left(  m\otimes
n\right)  }_{=m_{\left(  0\right)  }\otimes m_{\left(  1\right)  }S\left(
n_{\left(  -1\right)  }\right)  \otimes n_{\left(  0\right)  }}\right)
\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id}
\nolimits_{N}\right)  \left(  \underbrace{\left(  \operatorname*{id}
\nolimits_{M\otimes H}\otimes\rho_{\ell}\right)  \left(  m_{\left(  0\right)
}\otimes m_{\left(  1\right)  }S\left(  n_{\left(  -1\right)  }\right)
\otimes n_{\left(  0\right)  }\right)  }_{=m_{\left(  0\right)  }\otimes
m_{\left(  1\right)  }S\left(  n_{\left(  -1\right)  }\right)  \otimes
\rho_{\ell}\left(  n_{\left(  0\right)  }\right)  }\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id}
\nolimits_{N}\right)  \left(  m_{\left(  0\right)  }\otimes m_{\left(
1\right)  }S\left(  n_{\left(  -1\right)  }\right)  \otimes\underbrace{\rho
_{\ell}\left(  n_{\left(  0\right)  }\right)  }_{=\left(  n_{\left(  0\right)
}\right)  _{\left(  -1\right)  }\otimes\left(  n_{\left(  0\right)  }\right)
_{\left(  0\right)  }}\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id}
\nolimits_{N}\right)  \left(  m_{\left(  0\right)  }\otimes m_{\left(
1\right)  }S\left(  n_{\left(  -1\right)  }\right)  \otimes\left(  n_{\left(
0\right)  }\right)  _{\left(  -1\right)  }\otimes\left(  n_{\left(  0\right)
}\right)  _{\left(  0\right)  }\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id}
\nolimits_{N}\right)  \left(  m_{\left(  0\right)  }\otimes m_{\left(
1\right)  }S\left(  n_{\left(  -2\right)  }\right)  \otimes n_{\left(
-1\right)  }\otimes n_{\left(  0\right)  }\right)  $
$=m_{\left(  0\right)  }\otimes m_{\left(  1\right)  }\underbrace{S\left(
n_{\left(  -2\right)  }\right)  n_{\left(  -1\right)  }}_{=\varepsilon\left(
n_{\left(  -1\right)  }\right)  1}\otimes n_{\left(  0\right)  }$
$=m_{\left(  0\right)  }\otimes m_{\left(  1\right)  }\varepsilon\left(
n_{\left(  -1\right)  }\right)  1\otimes n_{\left(  0\right)  }
=\underbrace{m_{\left(  0\right)  }\otimes m_{\left(  1\right)  }}_{=\rho
_{r}\left(  m\right)  }\otimes\underbrace{\varepsilon\left(  n_{\left(
-1\right)  }\right)  n_{\left(  0\right)  }}_{=n}$
$=\rho_{r}\left(  m\right)  \otimes n=\left(  \rho_{r}\otimes
\operatorname*{id}\nolimits_{N}\right)  \left(  m\otimes n\right)  $.
This proves (2). Next, we claim that
(3) $\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes
\operatorname*{id}\nolimits_{N}\right)  \circ\left(  \operatorname*{id}
\nolimits_{M\otimes H}\otimes\rho_{\ell}\right)  \circ\Omega
=\operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}$
as $\mathbf{k}$-linear maps $M\otimes N\rightarrow M\otimes H\otimes N$.
Indeed, for any $m\in M$ and $n\in N$, we have
$\left(  \left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes
\operatorname*{id}\nolimits_{N}\right)  \circ\left(  \operatorname*{id}
\nolimits_{M\otimes H}\otimes\rho_{\ell}\right)  \circ\Omega\right)  \left(
m\otimes n\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id}
\nolimits_{N}\right)  \left(  \left(  \operatorname*{id}\nolimits_{M\otimes
H}\otimes\rho_{\ell}\right)  \left(  \underbrace{\Omega\left(  m\otimes
n\right)  }_{=m\otimes1\otimes n}\right)  \right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id}
\nolimits_{N}\right)  \left(  \underbrace{\left(  \operatorname*{id}
\nolimits_{M\otimes H}\otimes\rho_{\ell}\right)  \left(  m\otimes1\otimes
n\right)  }_{=m\otimes1\otimes\rho_{\ell}\left(  n\right)  }\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id}
\nolimits_{N}\right)  \left(  m\otimes1\otimes\underbrace{\rho_{\ell}\left(
n\right)  }_{=n_{\left(  -1\right)  }\otimes n_{\left(  0\right)  }}\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id}
\nolimits_{N}\right)  \left(  m\otimes1\otimes n_{\left(  -1\right)  }\otimes
n_{\left(  0\right)  }\right)  $
$=m\otimes\underbrace{1n_{\left(  -1\right)  }\otimes n_{\left(  0\right)  }
}_{=n_{\left(  -1\right)  }\otimes n_{\left(  0\right)  }=\rho_{\ell}\left(
n\right)  }=m\otimes\rho_{\ell}\left(  n\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}\right)  \left(
m\otimes n\right)  $.
This proves (3). Subtracting (3) from (2), we obtain
$\left(  \operatorname*{id}\nolimits_{M}\otimes\mu\otimes\operatorname*{id}
\nolimits_{N}\right)  \circ\left(  \operatorname*{id}\nolimits_{M\otimes
H}\otimes\rho_{\ell}\right)  \circ\left(  \Psi-\Omega\right)  =\rho_{r}
\otimes\operatorname*{id}\nolimits_{N}-\operatorname*{id}\nolimits_{M}
\otimes\rho_{\ell}$.
Hence,
(4) $\ker\left(  \Psi-\Omega\right)  \subseteq\ker\left(  \rho_{r}
\otimes\operatorname*{id}\nolimits_{N}-\operatorname*{id}\nolimits_{M}
\otimes\rho_{\ell}\right)  $.
On the other hand, let $\gamma:H\otimes N\rightarrow H\otimes N$ be the
$\mathbf{k}$-linear map defined by the requirement that
$\gamma\left(  h\otimes n\right)  =hS\left(  n_{\left(  -1\right)  }\right)
\otimes n_{\left(  0\right)  }$ for all $h\in H$ and $n\in N$.
(Avoiding Sweedler notation, we could define $\gamma$ as $\left(  \mu
\otimes\operatorname*{id}\nolimits_{N}\right)  \circ\left(  \operatorname*{id}
\nolimits_{H}\otimes S\otimes\operatorname*{id}\nolimits_{N}\right)
\circ\left(  \operatorname*{id}\nolimits_{H}\otimes\rho_{\ell}\right)  $, but
we wouldn't gain much from this definition.)
Now, I claim that
(5) $\left(  \operatorname*{id}\nolimits_{M}\otimes\gamma\right)
\circ\left(  \rho_{r}\otimes\operatorname*{id}\nolimits_{N}\right)  =\Psi$
as $\mathbf{k}$-linear maps $M\otimes N\rightarrow M\otimes H\otimes N$.
Indeed, for any $m\in M$ and $n\in N$, we have
$\left(  \left(  \operatorname*{id}\nolimits_{M}\otimes\gamma\right)
\circ\left(  \rho_{r}\otimes\operatorname*{id}\nolimits_{N}\right)  \right)
\left(  m\otimes n\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\gamma\right)  \left(
\underbrace{\left(  \rho_{r}\otimes\operatorname*{id}\nolimits_{N}\right)
\left(  m\otimes n\right)  }_{=\rho_{r}\left(  m\right)  \otimes n}\right)
=\left(  \operatorname*{id}\nolimits_{M}\otimes\gamma\right)  \left(
\underbrace{\rho_{r}\left(  m\right)  \otimes n}_{=m_{\left(  0\right)
}\otimes m_{\left(  1\right)  }\otimes n}\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\gamma\right)  \left(
m_{\left(  0\right)  }\otimes m_{\left(  1\right)  }\otimes n\right)
=m_{\left(  0\right)  }\otimes\underbrace{\gamma\left(  m_{\left(  1\right)
}\otimes n\right)  }_{\substack{=m_{\left(  1\right)  }S\left(  n_{\left(
-1\right)  }\right)  \otimes n_{\left(  0\right)  }\\\text{(by the definition
of }\gamma\text{)}}}$
$=m_{\left(  0\right)  }\otimes m_{\left(  1\right)  }S\left(  n_{\left(
-1\right)  }\right)  \otimes n_{\left(  0\right)  }=\Psi\left(  m\otimes
n\right)  $.
This proves (5). Furthermore, I claim that
(6) $\left(  \operatorname*{id}\nolimits_{M}\otimes\gamma\right)
\circ\left(  \operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}\right)
=\Omega$
as $\mathbf{k}$-linear maps $M\otimes N\rightarrow M\otimes H\otimes N$.
Indeed, for any $m\in M$ and $n\in N$, we have
$\left(  \left(  \operatorname*{id}\nolimits_{M}\otimes\gamma\right)
\circ\left(  \operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}\right)
\right)  \left(  m\otimes n\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\gamma\right)  \left(
\underbrace{\left(  \operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}\right)
\left(  m\otimes n\right)  }_{=m\otimes\rho_{\ell}\left(  n\right)  }\right)
=\left(  \operatorname*{id}\nolimits_{M}\otimes\gamma\right)  \left(
\underbrace{m\otimes\rho_{\ell}\left(  n\right)  }_{=m\otimes n_{\left(
-1\right)  }\otimes n_{\left(  0\right)  }}\right)  $
$=\left(  \operatorname*{id}\nolimits_{M}\otimes\gamma\right)  \left(
m\otimes n_{\left(  -1\right)  }\otimes n_{\left(  0\right)  }\right)
=m\otimes\underbrace{\gamma\left(  n_{\left(  -1\right)  }\otimes n_{\left(
0\right)  }\right)  }_{\substack{=n_{\left(  -1\right)  }S\left(  \left(
n_{\left(  0\right)  }\right)  _{\left(  -1\right)  }\right)  \otimes\left(
n_{\left(  0\right)  }\right)  _{\left(  0\right)  }\\\text{(by the definition
of }\gamma\text{)}}}$
$=m\otimes n_{\left(  -1\right)  }S\left(  \left(  n_{\left(  0\right)
}\right)  _{\left(  -1\right)  }\right)  \otimes\left(  n_{\left(  0\right)
}\right)  _{\left(  0\right)  }=m\otimes\underbrace{n_{\left(  -2\right)
}S\left(  n_{\left(  -1\right)  }\right)  }_{=\varepsilon\left(  n_{\left(
-1\right)  }\right)  1}\otimes n_{\left(  0\right)  }$
$=m\otimes\varepsilon\left(  n_{\left(  -1\right)  }\right)  1\otimes
n_{\left(  0\right)  }=m\otimes1\otimes\underbrace{\varepsilon\left(
n_{\left(  -1\right)  }\right)  n_{\left(  0\right)  }}_{=n}=m\otimes1\otimes
n=\Omega\left(  n\right)  $.
This proves (6). Subtracting (6) from (5), we obtain
(7) $\left(  \operatorname*{id}\nolimits_{M}\otimes\gamma\right)
\circ\left(  \rho_{r}\otimes\operatorname*{id}\nolimits_{N}-\operatorname*{id}
\nolimits_{M}\otimes\rho_{\ell}\right)  =\Psi-\Omega$.
Hence,
$\ker\left(  \rho_{r}\otimes\operatorname*{id}\nolimits_{N}-\operatorname*{id}
\nolimits_{M}\otimes\rho_{\ell}\right)  \subseteq\ker\left(  \Psi
-\Omega\right)  $.
Combining this with (4), we find
$\ker\left(  \rho_{r}\otimes\operatorname*{id}\nolimits_{N}-\operatorname*{id}
\nolimits_{M}\otimes\rho_{\ell}\right)  =\ker\left(  \Psi-\Omega\right)  $.
Now, the first definition of $M\square_{H}N$ yields
$M\square_{H}N=\ker\left(  \rho_{r}\otimes\operatorname*{id}\nolimits_{N}
-\operatorname*{id}\nolimits_{M}\otimes\rho_{\ell}\right)  =\ker\left(
\Psi-\Omega\right)  $
$=\ker\left(  \left(  \operatorname*{id}\nolimits_{M}\otimes\tau\right)
\circ\left(  \Psi-\Omega\right)  \right)  $
(since $\operatorname*{id}\nolimits_{M}\otimes\tau$ is a $\mathbf{k}$-module
isomorphism (since $\tau$ is a $\mathbf{k}$-module somorphism))
(8) $=\left\{  t\in M\otimes N\ \mid\ \left(  \left(  \operatorname*{id}
\nolimits_{M}\otimes\tau\right)  \circ\left(  \Psi-\Omega\right)  \right)
\left(  t\right)  =0\right\}  $.
But if $t\in M\otimes N$ is given as a sum of pure tensors as follows:
$t=\sum_{i}m_{i}\otimes n_{i}$,
then
$\left(  \left(  \operatorname*{id}\nolimits_{M}\otimes\tau\right)
\circ\left(  \Psi-\Omega\right)  \right)  \left(  t\right)  $
$=\sum_{i}\left(  m_{i}\right)  _{\left(  0\right)  }\otimes\left(
n_{i}\right)  _{\left(  0\right)  }\otimes\left(  m_{i}\right)  _{\left(
-1\right)  }S\left(  \left(  n_{i}\right)  _{\left(  1\right)  }\right)
-\sum_{i}m_{i}\otimes n_{i}\otimes1$,
and therefore (8) rewrites as follows:
$M\square_{H}N$
$=\left\{  \sum_{i}m_{i}\otimes n_{i}\in M\otimes N\ \mid\ \sum_{i}\left(
m_{i}\right)  _{\left(  0\right)  }\otimes\left(  n_{i}\right)  _{\left(
0\right)  }\otimes\left(  m_{i}\right)  _{\left(  -1\right)  }S\left(  \left(
n_{i}\right)  _{\left(  1\right)  }\right)  =\sum_{i}m_{i}\otimes n_{i}
\otimes1\right\}  $.
