Tensor product of $R$-algebras Let $f: R \to S$ and $g: R \to T$ be two $R$-algebras. To show that $S \otimes_R T$ is an $R$-algebra I need to define a ring structure (multiplication) on it and a ring homomorphism $h : R \to S \otimes_R T$.
Using the universal property of the (multi-)tensor product, defining multiplication is clear to me. What I'm confused about is the map $h : R \to S \otimes_R T$. According to this answer here, $r \mapsto 1 \otimes g(r) = f(r) \otimes 1$. 
How are they the same?
 A: The basic fact is that for  two $R$-modules $M,N$ you have:
 $$r\cdot m\otimes _R n=m\otimes_R r\cdot n \quad (\text {for all} \quad  r\in R,\; m\in M,\; n\in N)  \quad (*)$$
In particular, in your case, you have:  $$r\cdot 1_S\otimes _R 1_T=1_S \otimes_R r\cdot 1_T \quad (**)$$   
In order to conclude, you just have to remember that built into the notion of algebra is the equality $r\cdot s=f(r)s$ where on the right hand side of the equality $f(r)s$ means the product of the elements $f(r),s$ in the ring $S$.
In particular $r\cdot 1_S=f(r)1_S=f(r)$ and similarly $r\cdot 1_T=g(r)$.
Transporting these last equalities into $(**)$, you get the required equation 
$$       f(r) \otimes_R 1_T=  1_S \otimes_R g(r)        \quad (***) $$ 
A: I believe your ring $R$ is supposed to be commutative, otherwise see here for example https://mathoverflow.net/questions/21899/definition-of-an-algebra-over-a-noncommutative-ring. 
The answer is hidden in the what are the actions of $R$ on the two algebras. An $R$-algebra is a ring with an action of $R$ (which is a left and right action since $R$ is commutative), and the maps $f \colon R \to S$ and $f \colon R \to T$ give you the actions. 
They are $R \times S \to S \colon (r,s) \mapsto r \ast s := f(r) s$, with the multiplication of $S$, and similarly for $T$.
Therefore, in the tensor product $S \otimes_{R} T$, the left action on $T$ and the right action on $S$ are needed in order to define the tensor product. Moreover, the left action on $S$ and the right action on $T$ will give the additional structure of $R$-algebra, and these are the ones we use.
To come back to your question, the equality is proved by
$f(r) \otimes 1_T = (f(r) \cdot 1_{S}) \otimes 1_T = (r \ast 1_S) \otimes 1_T = 1_S \otimes (1_T \ast r) = 1_S \otimes (1_T \cdot g(r)) = 1_S \otimes g(r)$, where the essential step is the use of the relation $r a \otimes b = a \otimes  br$.
