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I'm reading this and am trying to unpack the definition of $Alg_2(k)$ found at the bottom of page 21 so I'm wondering if the following is a good example of what's being described.

Let $R= M_{n\times n}(\mathbb{R})$ where $M_{n\times n}(\mathbb{R})$ is the ring of $n\times n$ matrices. Now let $R$ be a module by considering it as a module over itself. Similary, let $S$ be the module of $m\times m$ matrices. Now, it's easy to show that $M_{n\times m}(\mathbb{R})$ is an $R$-$S$ bimodule. We also have the following map. Take $A\in M_{n\times m}(\mathbb{R})$, then $A$ defines a map from $ M_{n\times n}(\mathbb{R})\to M_{m\times m}(\mathbb{R})$ by $x\mapsto A^Tx A$.

My question is, is the map $x\mapsto A^Tx A$ an element of $\operatorname{Maps}_{Alg_2(k)}(R,S)$?

Edit: Fixed some typos and tried to make my notation more clear. Also, other examples of elements of $\operatorname{Maps}_{Alg_2(k)}(R,S)$ are welcome.

Edit2: Incorporated notation suggestions.

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    This is full of typos, and not all of your notation is defined. You might give it a quick proofread. – 2017-02-27
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    @KevinCarlson Fair enough. Since my example is lacking I'll see if I can clean it up but also ask for other examples. – 2017-02-27
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    "$A'nA$ takes $n\in M_{n\times n}(\Bbb R)$ to $s\in M_{m\times m}(\Bbb R)$" makes no sense. (Also you mean to call $S$ a ring, not a module, I assume.) – 2017-02-27
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    @arctictern Hmm, all I mean is $(m\times n)(n\times n)(n\times m)$ gives you a a square $(m\times m)$ matrix. I meant to consider $S$ as a module over itself. – 2017-02-27
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    You seem to be using $S$ as a ring, and not as a module over itself. When you write, $n\in M_{n\times n}(\Bbb R)$, you're saying $n$ is both an integer and a matrix. And you never said what $s$ is. You seem to be defining $s$ to be $A^TxA$ where $x\in M_{n\times n}(\Bbb R)$. Then you should just say every matrix $A\in M_{n\times m}(\Bbb R)$ defines a map $M_{n\times n}(\Bbb R)\to M_{m\times m}(\Bbb R)$ given by $x\mapsto A^TxA$. In other words, a vector space map $R\to S$. – 2017-02-27
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    But a vector space map $R\to S$ isn't really relevant to us. As the pdf says, ${\rm Maps}_{{\rm Alg}_2(k)}(R,S)$ represents a collection of functors from the category of $R$-modules to the category of $S$-modules, and I don't see how defining an $\mathbb{R}$-linear map $R\to S$ is supposed to define such a functor between categories of modules. – 2017-02-27
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    @arctictern Well, I'm trying to use the idea that an R-S bimodule can be taken to be a functor between R-modules and S-modules and produce a tangible example. – 2017-02-27

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Have a look at the nLab article; it should clear up any confusion.

Briefly, the 2-category they are describing has $k$-algebras as the objects, and a 1-morphism from a $k$-algebra $A$ to a $k$-algebra $B$ is an $(A,B)$-bimodule $V$. Bimodules are composed using the tensor product. So given a $(B,C)$-bimodule $W$, you can compose with $V$ by tensoring over $B.$ The $(A,C)$-bimodule $V \otimes_B W$ is the composite of $V$ and $W$. Thus, the identity 1-morphism on a $k$-algebra $A$ is just $A$ itself viewed as an $A$-bialgebra. Indeed, tensoring $A \otimes_A V \cong V.$

Since bimodules are the 1-morphisms, 2-morphisms must take us between bimodules, and the obvious notion here is the correct one: 2-morphisms are just bimodule homomorphisms. The vertical composition is just the usual composition of bimodule homomorphisms.

If it still doesn't make sense to you, you probably need to review tensor products and bimodules in your favourite abstract algebra text (such as Aluffi, for example).

Note: I have use the opposite of the usual composition convention for the 1-morphisms, because it is easier to follow in this situation.