what are the arrows in between the objects $M \otimes_U N$ in the category $\mathscr{M} \otimes_U N$? I am trying to understand a single object category. May be we can call it as monoid category.
Suppose $R$ be a commutative ring with identity and $A$ is a $R$-algebra.

$(1)$ Can we think of $A$ as a category with a single object $A$ itself and with the identity arrow
?

For better sense we can talk about monoidal category with morphisms $A \otimes A \to A$ and identity map $id \to A$  with a commutative diagram.
Actually I am trying to understand the product between a category of modules over a commutative ring and a module over another commutative ring. More specifically, let $U$ and $V$ be two commutative rings with  $U \subset V$. Let $\mathscr{M}$ be the category of $V$-modules and $N$ be a $U$-algebra.
Now I want to define the product $\mathscr{M} \otimes_U N$  or just $\mathscr{M} \times N$.  I have looked the definition $4.2.20$ in The Stack Project.
I understand that the objects of the new category $\mathscr{M} \otimes_U N$ will be of the form $M \otimes_U N$ or just $M \times N$ for $M \in \mathscr{M}$.

$(2)$ But what are the arrows in between the objects $M \otimes_U N$ in $\mathscr{M} \otimes_U N$ ?

If the question $(1)$ is true, then I think the arrows  or morphisms in the new category $\mathscr{M} \otimes_U N$ are just $f \otimes 1$, where $f$ is the arrow in the category $\mathscr{M}$ and $1$ denote the identity arrow in the single object category $N$.
Any comment on question $(1)$ and question $(2)$
 A: Let me answer these questions in order.
First, let $R$ be any ring with identity. In general we can regard $R$ as a category with one object, $\mathscr{R}$, defined as follows:
Objects: A singleton set $\lbrace \ast \rbrace$.
Morphisms: Define the hom-set $\mathscr{R}(\ast,\ast):= R$.
Composition: For any $r,s \in R$, define $r \circ s := rs$, i.e., the composition of endomorphisms of $\ast$ is given by multiplication in the ring $R$.
Identities: The identity on $\ast$ is $1_R$.
As you anticipated, this does become a monoidal category: $\mathscr{R}$ has monoidal functor $\otimes:\mathscr{R} \times \mathscr{R} \to \mathscr{R}$ given by sending $(\ast,\ast) \mapsto \ast$ and sending a pair of morphisms $(r,s) \mapsto r \circ s = rs$.  This actually gives a functor $\mathbf{Ring} \to \mathbf{MonCat}$, from the $1$-category of rings to the $1$-category of monoidal categories with monoidal functors between them.
Now assume that $R$ is commutative. If $A$ is a unital $R$-algebra, then following the process above we exactly get that $A$ may be represented by a one-object category $\mathscr{A}$ (just proceed mutatis mutandis as to how we built $\mathscr{R}$). Things get really annoying if we have non-unital algebras, but we can always use the unitization functor $R\mathbf{Alg} \to R\mathbf{Alg}_1$, tensor with $R$ appropriately, and then make sure to keep track of augmentation ideals appropriately. Because of this, I'm going to focus on unital $R$-algebras.
An interesting and important observation of this unital ring <--> $\mathbf{Ab}$-enriched monoidal category with one object is that an action of the category $\mathscr{R}$ is on $\mathscr{A}$ which respects all the relevant monoidal structures (an actegorical structure) is equivalent to making sure that $\mathscr{R}$ acts appropriately on the set of endomorphisms of the identity functor, $Z(\mathscr{A}) = [id_{\mathscr{A}},id_{\mathscr{A}}]$. Because the category $\mathscr{A}$ is enriched over the category of Abelian groups, $\mathbf{Ab}$, and this action is bilinear with respect to composition in $\mathscr{A}$, it isn't too rough to show that $Z(\mathscr{A})$ is a (commutative) unital ring. However, $Z(\mathscr{A}) \cong Z(A)$ as rings, so such an action corresponds to a ring morphism $R \to A$ which factors through $Z(A)$, i.e., this corresponds to how we witness $A$ as an $R$-algebra.
In this way we answer your question (1) affirmatively: Up to some isomorphism and unitization issues that I brushed under the rug, we can indeed regard an algebra $A$ as a one-object category (equipped with an appropriate action of the one-object category of $R$, of course).
Your second question is a little more involved. Following your notation, fix two commutative unital rings $U, V$ with an inclusion ring morphism $U \hookrightarrow V$. Let $V\mathbf{Mod}$ be the category of $V$-modules and let $N$ be a $U$-algebra. As far as I have seen, the category $V\mathbf{Mod} \otimes_U N$ is defined as follows (although your mileage may vary based on context of where you have seen this category):
Objects: $U$-modules $M \otimes_U N$, where we regard $M$ as a $U$-module via restriction of scalars.
Morphisms: $U$-linear morphisms $f \otimes g:M \otimes_U N \to M^{\prime} \otimes_U N$ where $f \in V\mathbf{Mod}(M,M^{\prime})$ and $g \in U\mathbf{Mod}(N,N)$.
Composition: Assume that $f \otimes g:M \otimes_U N \to M^{\prime} \otimes_U N$ and $h \otimes k:M^{\prime} \otimes_U N \to M^{\prime\prime} \otimes_U N$ are morphisms in $V\mathbf{Mod} \otimes_U N$. The composite
$$
(h \otimes k) \circ (f \otimes g) := (h \circ f) \otimes (k \circ g):M \otimes_U N \to M^{\prime\prime} \otimes_U N.
$$
Identities: For any object $M \otimes_U N$ in $V\mathbf{Mod} \otimes_U N$, the identity morphism $id_{M \otimes_U N}$ is $id_M \otimes id_N$.
Now if you write $\mathscr{N}$ as a single object category, there is a connection between
$$
V\mathbf{Mod} \otimes_U N
$$
and
$$
V\mathbf{Mod} \otimes_{\mathscr{U}} \mathscr{N},
$$
where this second category has some horrible abuse of notation going on. By the second category, I mean the category with objects $(M,\ast)$, for a $V$-module $M$ regarded as a $U$-module, and $\ast$ the object of $\mathscr{N}$ (you think of this being the object $M \otimes_U N$). A morphism in $V\mathbf{Mod} \otimes_{\mathscr{U}} \mathscr{N}$ corresponds to a pair $(f,g)$ where $f$ is a morphism in $V\mathbf{Mod}$ and $g$ is a $U$-module endomorphism of $N$ induced by multiplication by an element of $N$ (so by, for instance, the left-multiplication-by $n \in N$ map). Composition is given component-wise, and the identities are $(id_M,id_{\ast}) = (id_M,1_N)$.
Generally there is an inclusion $V\mathbf{Mod} \otimes_{\mathscr{U}} \mathscr{N} \to V\mathbf{Mod} \otimes_U N$, but it need not be an isomorphism unless $U\mathbf{Mod}(N,N) \cong N$. It is also not true that $V\mathbf{Mod} \times \mathscr{N}$ is equivalent to $V\mathbf{Mod} \otimes_U N$ or $V\mathbf{Mod} \otimes_{\mathscr{U}} \mathscr{N}$. The issue here is that in the product category $V\mathbf{Mod} \times \mathscr{N}$, there is no restriction of scalars to $U$ business. Objects are just $(M,\ast)$ where $M$ is a $V$-module. You still need to somehow be able to collapse things via a restriction of scalars argument, and in general you can have two non-isomorphic $V$-modules that become isomorphic after restricting scalars to a smaller ring $U$.
For an example of this second phenomenon, take $U = \mathbb{Q}, V = \mathbb{Q}[x]$, and let the map $U \to V$ the standard embedding. Now note that a $\mathbb{Q}[x]$-module is equivalent to a $\mathbb{Q}$-vector space $V$ equipped with a $\mathbb{Q}$-linear endomorphism $T:V \to V$.  Choose the $\mathbb{Q}[x]$-modules as follows: Let $M = (\mathbb{Q}^2,T)$ where $T$ is a linear transformation that acts via the matrix
$$
\begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}
$$
in the standard basis for $\mathbb{Q}^2$ and let $M^{\prime}$ be given by the underlying vector space $\mathbb{Q}^2$ with linear transformation $T^{\prime}$ acting via the matrix
$$
\begin{pmatrix}
0 & 1 \\ 0 & 0
\end{pmatrix}
$$
in the standard basis of $\mathbb{Q}^2$. Then it is routine to see that in $\mathbb{Q}[x]\mathbf{Vect}$, $M \not\cong M^{\prime}$ (one linear transformation is the identity while the other is nilpotent). However, restricting scalars from $\mathbb{Q}[x]$ to $\mathbb{Q}$ gives that
$$
\operatorname{Res}(M) = \operatorname{Res}(\mathbb{Q}^2,T) = \mathbb{Q}^2 = \operatorname{Res}(\mathbb{Q}^2,T^{\prime}) = \operatorname{Res}(M^{\prime}).
$$
which shows the phenomenon.
As a final comment to this alreadly way too long answer, in order to get a category with where objects are pairs $(M,\ast)$ and where morphisms are $(f,id_{\ast})$ for a one-object category $\mathscr{N}$ and any category $\mathscr{M}$ with $M \in \mathscr{M}_0$ and $f \in \mathscr{M}_1$, then you need the category
$$
\mathscr{M} \times \underline{1},
$$
where $\underline{1}$ is the category with a single object and single arrow. Unsurprisingly, this category is canonically isomorphic to $\mathscr{M}$.
