About the closed structure on the modules of a monoidal closed symmetrical  category  Let $(\mathscr{C}, \otimes , I)$ monoidal category, a  monoid $(R, e_R, \mu )$ is a object $R \in  \mathscr{C}$ with morphisms $e_R: I \to  R$,  $\mu: R  \otimes  R \to  R$ with the well knowed unital e associative commutative diagram.
If  $(\mathscr{C}, \otimes , I)$ has a symmetry (is enought a braid) then the tensor prodoct $M \otimes   N $ of two objects, each one with a monoid structure, has a natural monoid structure, then we have a monoidal symmetrical (braided) category of monoid of $(\mathscr{C}, \otimes , I)$. Given a monoid $R$ a  left $R$-module $(M, \mu)$ is a object $m \in  C$ with a morphism $\mu: R \otimes   M \to  M $ with usual unitary and associative diagram, similarly are defined the right modules.
Now we suppose that $(\mathscr{C}, \otimes, [-, ?],  I)$ ia  a monoidal closed, symmetrical category, given a right $R$-module $M$and a a left $R$-module $N$,  In analogy to the classical algebra context, let $[M, N]^R $ the kernel of the two morphisms gived by the compositions:
$[M, N] \xrightarrow{- \otimes  R} [M \otimes   R, N \otimes   R ] \xrightarrow{[1, \cong ]} [M \otimes  R , R \otimes N ] \xrightarrow{[1, \mu_N]}[R \otimes  M, N ]$
$[M, N] \xrightarrow{ [\mu_M, 1] } [R \otimes  M, N ]$
How to prove that $[M, N]^R$ is a  left $R$-module (for the structure inducted  by $M$, and inducted by $N$) ?
This is mentioned on §3 of the Stefan Schwede & Brooke Shipley  article 'Algebras and modules
in monoidal model categories'
http://arxiv.org/abs/math/9801082
I know that A. Kock studied the monoidal -closed structure of algebras of a (commutative) triple in monoidal closed symmetrical categories (then a much more general and complex result).
I ask where I can find a explicit but rigorous proof.  
 A: I take it you are referring to pages 3-4 of the cited paper. Please note that there $R$ is taken to be a commutative monoid, not just a monoid as in your post. 
I'm not sure where a proof is written out, but the main message is that you can do this easily enough yourself. Just keep in mind that the proof you want is not essentially different from the familiar one for modules over a commutative ring $R$, so you should probably right out the details of that proof first, using good old-fashioned elements, and straightforwardly translate that proof into commutative diagrams. For example, the equalizer you mentioned, when interpreted in the symmetric monoidal closed category of abelian groups, just expresses the following subgroup of $Hom_{Ab}(M, N)$: 
$$[M, N]^R := \{M \stackrel{f}{\to} N:  \forall r \in R, m \in M \ (f(rm) = rf(m))\},$$ 
in other words the subgroup of $R$-module maps. We are trying to show that this carries an $R$-module structure if $R$ is commutative. 
The $R$-action will of course be the pointwise one. In the language of elements, if $s \in R$ and $f: M \to N$, we define $s f$ by the rule $(sf)(m) := s(f(m))$. In terms of diagrams, 
the desired module structure on $[M, N]$, $R \otimes [M, N] \to [M, N]$, corresponds to the composite 
$$R \otimes [M, N] \otimes M \stackrel{eval_{M, N}}{\to} R \otimes N \stackrel{\alpha}{\to} N$$ 
where $\alpha$ is the $R$-module structure on $N$. 
We want to show this $R$-action restricts to an $R$-action $R \otimes [M, N]^R \to [M, N]^R$. In the language of elements, we thus want to show that if $f$ is an $R$-module map, then so is $r f$ for each $r \in R$, i.e., for all $s \in R$ and $m \in M$, we have 
$$(r f)(s m) = s((rf)(m)).$$ 
This is straightforward: 
$$(r f)(s m) \stackrel{def}{=} r(f (s m)) = r(s f(m)) = (r s)(f(m)) = (s r)(f(m)) = s(r f(m)) \stackrel{def}{=} s((rf)(m))$$ 
Now all you have to do is translate these equations into commutative diagrams. It's a little tedious, but not difficult. Bon courage. 
Edit (11/28/12): In case the answer given above was not completely satisfactory, I have written out full details which can be found here. 
