[proofplan]
For part (1), existence is established by defining an $R$-bilinear map $A \times B \to C$ via $f_1$ and $f_2$, using the universal property of $A \otimes_R B$ as an $R$-module to lift it to an $R$-module homomorphism $h$, and then verifying $h$ is multiplicative and unital. Uniqueness follows because the pure tensors $\{a \otimes 1\}$ and $\{1 \otimes b\}$ generate $A \otimes_R B$ as an $R$-algebra, so any $R$-algebra homomorphism is determined by its values on these generators. Part (2) is the standard uniqueness-of-universal-objects argument: apply (1) in both directions to obtain mutually inverse $R$-algebra homomorphisms.
[/proofplan]
[step:Prove uniqueness of $h$ in part (1)]
Every element of $A \otimes_R B$ is an $R$-linear combination of pure tensors $a \otimes b$. Each pure tensor factors as
\begin{align*}
a \otimes b = (a \otimes 1)(1 \otimes b) = i_A(a) \cdot i_B(b),
\end{align*}
since the multiplication in $A \otimes_R B$ satisfies $(a \otimes 1)(1 \otimes b) = a \cdot 1 \otimes 1 \cdot b = a \otimes b$. Therefore $A \otimes_R B$ is generated as an $R$-algebra by $i_A(A) \cup i_B(B)$.
If $h: A \otimes_R B \to C$ is any $R$-algebra homomorphism with $h \circ i_A = f_1$ and $h \circ i_B = f_2$, then on pure tensors:
\begin{align*}
h(a \otimes b) = h(i_A(a) \cdot i_B(b)) = h(i_A(a)) \cdot h(i_B(b)) = f_1(a) \cdot f_2(b).
\end{align*}
Since $h$ is $R$-linear and pure tensors span $A \otimes_R B$ as an $R$-module, $h$ is uniquely determined.
[/step]
[step:Construct $h$ via the universal property of $A \otimes_R B$ as an $R$-module]
Define the map $\Phi: A \times B \to C$ by $\Phi(a, b) = f_1(a) \cdot f_2(b)$. We verify that $\Phi$ is $R$-bilinear. For $a, a' \in A$, $b \in B$, and $r \in R$:
\begin{align*}
\Phi(a + a', b) &= f_1(a + a') f_2(b) = (f_1(a) + f_1(a')) f_2(b) = f_1(a) f_2(b) + f_1(a') f_2(b) = \Phi(a, b) + \Phi(a', b), \\
\Phi(ra, b) &= f_1(ra) f_2(b) = r f_1(a) f_2(b) = r \Phi(a, b), \\
\Phi(a, rb) &= f_1(a) f_2(rb) = f_1(a) \cdot r f_2(b) = r f_1(a) f_2(b) = r \Phi(a, b),
\end{align*}
where the second and third lines use the fact that $f_1$ and $f_2$ are $R$-algebra homomorphisms, so $f_1(ra) = rf_1(a)$ and $f_2(rb) = rf_2(b)$, and $r$ acts centrally in the commutative ring $C$. Bilinearity in the second variable follows by symmetry.
By the universal property of the tensor product $A \otimes_R B$ (as an $R$-module), there exists a unique $R$-module homomorphism
\begin{align*}
h: A \otimes_R B &\to C, \quad a \otimes b \mapsto f_1(a) f_2(b).
\end{align*}
[/step]
[step:Verify that $h$ is an $R$-algebra homomorphism]
**Unit.** Since $f_1$ and $f_2$ are $R$-algebra homomorphisms, $f_1(1_A) = 1_C$ and $f_2(1_B) = 1_C$. Therefore
\begin{align*}
h(1_A \otimes 1_B) = f_1(1_A) f_2(1_B) = 1_C \cdot 1_C = 1_C.
\end{align*}
**Multiplicativity.** It suffices to check on pure tensors, since $h$ is $R$-linear and pure tensors span $A \otimes_R B$. For pure tensors $a \otimes b$ and $a' \otimes b'$:
\begin{align*}
h((a \otimes b)(a' \otimes b')) &= h(aa' \otimes bb') = f_1(aa') f_2(bb') = f_1(a) f_1(a') f_2(b) f_2(b'),
\end{align*}
using the multiplicativity of $f_1$ and $f_2$. On the other hand:
\begin{align*}
h(a \otimes b) \cdot h(a' \otimes b') = f_1(a) f_2(b) \cdot f_1(a') f_2(b') = f_1(a) f_1(a') f_2(b) f_2(b'),
\end{align*}
where the last equality uses commutativity of $C$ to interchange $f_2(b)$ and $f_1(a')$. The two expressions agree, so $h$ is multiplicative on pure tensors. Since every element of $A \otimes_R B$ is a finite sum of pure tensors, multiplicativity extends by the distributive law.
[guided]
Why is commutativity of $C$ essential here? In the computation above, we needed $f_2(b) \cdot f_1(a') = f_1(a') \cdot f_2(b)$ to rearrange the product. If $C$ were non-commutative, the images of $A$ and $B$ in $C$ would need to commute with each other for the argument to work. This is precisely why the universal property of the tensor product of commutative algebras is stated for commutative target algebras $C$.
Let us also verify that the extension from pure tensors to all elements is legitimate. An arbitrary element $x \in A \otimes_R B$ has the form $x = \sum_{i=1}^m a_i \otimes b_i$. For two such elements $x = \sum_i a_i \otimes b_i$ and $y = \sum_j a_j' \otimes b_j'$:
\begin{align*}
h(xy) &= h\!\left(\sum_{i,j} a_i a_j' \otimes b_i b_j'\right) = \sum_{i,j} f_1(a_i a_j') f_2(b_i b_j') \\
&= \sum_{i,j} f_1(a_i) f_1(a_j') f_2(b_i) f_2(b_j') = \left(\sum_i f_1(a_i) f_2(b_i)\right)\left(\sum_j f_1(a_j') f_2(b_j')\right) = h(x) h(y),
\end{align*}
where the penultimate equality uses commutativity of $C$ and the distributive law.
[/guided]
**Compatibility with $i_A$ and $i_B$.** For all $a \in A$: $h(i_A(a)) = h(a \otimes 1_B) = f_1(a) f_2(1_B) = f_1(a) \cdot 1_C = f_1(a)$. For all $b \in B$: $h(i_B(b)) = h(1_A \otimes b) = f_1(1_A) f_2(b) = 1_C \cdot f_2(b) = f_2(b)$. Hence $h \circ i_A = f_1$ and $h \circ i_B = f_2$.
[/step]
[step:Prove part (2) by the standard uniqueness argument for universal objects]
Suppose $(Q, j_A, j_B)$ also satisfies the universal property of (1). Applying the universal property of $(A \otimes_R B, i_A, i_B)$ to the maps $j_A: A \to Q$ and $j_B: B \to Q$ yields a unique $R$-algebra homomorphism
\begin{align*}
\varphi: A \otimes_R B \to Q \quad \text{with} \quad \varphi \circ i_A = j_A, \quad \varphi \circ i_B = j_B.
\end{align*}
Applying the universal property of $(Q, j_A, j_B)$ to the maps $i_A: A \to A \otimes_R B$ and $i_B: B \to A \otimes_R B$ yields a unique $R$-algebra homomorphism
\begin{align*}
\psi: Q \to A \otimes_R B \quad \text{with} \quad \psi \circ j_A = i_A, \quad \psi \circ j_B = i_B.
\end{align*}
The composition $\psi \circ \varphi: A \otimes_R B \to A \otimes_R B$ satisfies $(\psi \circ \varphi) \circ i_A = \psi \circ j_A = i_A$ and $(\psi \circ \varphi) \circ i_B = \psi \circ j_B = i_B$. The identity map $\operatorname{id}_{A \otimes_R B}$ also satisfies these two conditions. By the uniqueness assertion in part (1) applied to the pair $(i_A, i_B)$ with target $A \otimes_R B$, we conclude $\psi \circ \varphi = \operatorname{id}_{A \otimes_R B}$.
By a symmetric argument, applying the uniqueness assertion of the universal property of $(Q, j_A, j_B)$ to the pair $(j_A, j_B)$ with target $Q$, we obtain $\varphi \circ \psi = \operatorname{id}_Q$.
Therefore $\varphi$ is an $R$-algebra isomorphism with inverse $\psi$. Uniqueness of $\varphi$ follows from the uniqueness in part (1).
[/step]
