Examples proving why the tensor product does not distribute over direct products.

We consider $\mathbb{Z}$-modules (i.e., abelian groups).

Since $\mathbb{Q}$ is divisible, if $A$ is a torsion abelian group, then $A\otimes\mathbb{Q}$ is trivial.

Let $G$ be the direct product of cyclic group of order $p^n$, with $p$ a prime, and $n$ increasing; that is: $$G = \prod_{n=1}^{\infty}\mathbb{Z}/p^n\mathbb{Z}.$$

Then $$\prod_{n=1}^{\infty}\left(\mathbb{Z}/p^n\mathbb{Z}\otimes\mathbb{Q}\right) = 0.$$

But $G\otimes\mathbb{Q}$ is not trivial: if we let $x$ be the element that corresponds to the class of $1$ in every coordinate, then $x$ has infinite order. Therefore, $$\langle x\rangle \otimes\mathbb{Q}\cong \mathbb{Z}\otimes\mathbb{Q} \cong\mathbb{Q};$$ but tensoring with $\mathbb{Q}$ over $\mathbb{Z}$ is exact; therefore, the embedding $\langle x\rangle \hookrightarrow G$ induces an embedding $\langle x\rangle\otimes \mathbb{Q}\hookrightarrow G\otimes \mathbb{Q}$. Therefore, $G\otimes\mathbb{Q}\neq 0$. Thus, we have $$\left(\prod_{n=1}^{\infty}\mathbb{Z}/p^n\mathbb{Z}\right)\otimes \mathbb{Q}\not\cong \prod_{n=1}^{\infty}(\mathbb{Z}/p^n\mathbb{Z}\otimes\mathbb{Q}).$$

Let $X$ and $Y$ be indeterminates.

For an example where your map is not surjective, take $M_i:=R$ for all $i\in\mathbb N$, and $N:=R[Y]$.

Then you get the natural map $$ R[[X]][Y]\to R[Y][[X]], $$ and $$ \sum_{i\in\mathbb N}\ X^i\ Y^i $$ is not in the image.

EDIT. Same example with different notation: Put $$ A:=\left(\prod M_i\right)\otimes N,\quad B:=\prod\ (M_i\otimes N), $$ and, for all $i,j\in\mathbb N$, $$ M_i=R_i=R_j=R_{ij}=R. $$ Set also $N:=\bigoplus R_j$. Then we have canonical isomorphisms $$ A=\bigoplus_j\ \prod_i\ R_{ij},\quad B=\prod_i\ \bigoplus_j\ R_{ij}. $$ We also have the inclusions $$ A\subset B\subset\prod_{i,j}\ R_{ij}, $$ and your map becomes the first inclusion.

Note that the Kronecker symbol $(\delta_{ij})$ is in $B$ but not in $A$.