Why is $B\otimes_A C$ a ring, where $B,C$ are $A$-algebra Given $B,C$ be two $A$-algebra, it is said that $B\otimes_AC$ has a ring structure, with the multiplication being defined as:
$$
(b_1\otimes c_1)\cdot(b_2\otimes c_2):=(b_1b_2)\otimes (c_1c_2).
$$
However I don't see an easy way to check it is well-defined.
For, given another pair of representatives:
$$
(b_1'\otimes c_1')\cdot(b_2'\otimes c_2'):=(b_1'b_2')\otimes (c_1'c_2')
$$
where
$$
b_1\otimes c_1=b_1'\otimes c_1'\quad\text{and}\quad b_2\otimes c_2=b_2'\otimes c_2'.
$$
How to verify that
$$
(b_1b_2)\otimes (c_1c_2)=(b_1'b_2')\otimes (c_1'c_2')?
$$

That $B$ is an $A$-algebra means that it comes equipped with maps $u_B : A \to B$ (unit) and $m_B : B \otimes_A B \to B$ (multiplication) of $A$-modules such that certain diagrams commute (write them down!). Similarly for $C$. But then $B \otimes_A C$ has the following $A$-algebra structure:
The unit is $A \cong A \otimes_A A \xrightarrow{u_B \otimes u_C} B \otimes_A C$. The multiplication is
$(B \otimes_A C) \otimes_A (B \otimes_A C) \cong (B \otimes_A B) \otimes_A (C \otimes_A C) \xrightarrow{m_B \otimes m_C} B \otimes_A C.$
So it is just induced by functoriality of the tensor product, as well as the usual associativity and symmetry isomorphisms. In fact, this procedure works in any symmetric monoidal category. No need to calculate with elements here.
