Theorem IV.$5.8$ of Hungerford's Algebra:
If $R$ and $S$ are rings and $A_R$, ${}_{R}B_{S}$, and ${}_{S}C$ are (bi)modules, then there is an isomorphism $$(A \otimes_RB)\otimes_S C \cong A \otimes_R(B\otimes_S C). $$
In the proof, the author says "verify that the assignment $(\sum_{i=1}^n a_i \otimes b_i,c)\mapsto \sum_{i=1}^n[a_i \otimes (b_i \otimes c_i)]$" defines an $S$-middle linear map $(A \otimes_RB) \times C \to A \otimes_R(B \otimes_S C)$.
I have a problem to prove this map is well-defined. Even, I cannot show $a_1 \otimes b_1=a_2 \otimes b_2$ implies $a_1 \otimes (b_1 \otimes c)=a_2 \times (b_2 \otimes c)$. How can I show that?
I want to define a map $\varphi \colon (A\otimes_R B) \otimes_S C \to A \otimes_R (B \otimes_S C)$.
To do this, I need to define an $S$-balanced map $\varphi' \colon (A\otimes_R B) \times C \to A \otimes_R (B \otimes_S C)$.
To do that, I'll first define an $R$-balanced map $\varphi''_c\colon A \times B \to A \otimes_R (B \otimes_S C)$ for each $c \in C$, given by $$\varphi''_c(a, b) := a \otimes_R(b\otimes_S c).$$
This is $R$-balanced because it is additive and $$ \varphi_c''(ar, b) = ar \otimes_R (b \otimes_S c) = a \otimes_R r(b \otimes_S c) = a \otimes_R (rb \otimes_S c) = \varphi''_c(a, rb). $$
The universal property for tensors then gives that there is a well-defined map $\widetilde{\varphi''_c}\colon A \otimes_R B \to A \otimes_R (B \otimes_S C)$ with $\widetilde{\varphi''_c}(a \otimes b) := \varphi''_c(a, b)$, and this holds for all $c$. Now we can go on to define:
$$ \varphi'(t, c) := \widetilde{\varphi''_c}(t). $$
To show that this is $S$-balanced, we need to show that $\widetilde{\varphi''_c}(ts) = \widetilde{\varphi''_{sc}}(t)$. Write $t = a_1 \otimes_R b_1 + \cdots + a_n \otimes_R b_n$. Then indeed, \begin{align*} \widetilde{\varphi''_c}(ts) &= \widetilde{\varphi''_c}(a_1 \otimes_R b_1s + \cdots + a_n \otimes_R b_ns) \\&= \widetilde{\varphi''_c}(a_1 \otimes_R b_1s) + \cdots + \widetilde{\varphi''_c}(a_n \otimes_R b_ns) \\&= \varphi''_c(a_1, b_1s) + \cdots + \varphi''_c(a_n, b_ns) \\&= a_1 \otimes_R (b_1s \otimes_S c) + \cdots + a_1 \otimes_R (b_1s \otimes_S c) \\&= a_1 \otimes_R (b_1 \otimes_S sc) + \cdots + a_1 \otimes_R (b_1 \otimes_S sc) \\&= \varphi''_{sc}(a_1, b_1) + \cdots + \varphi''_{sc}(a_n, b_n) \\&= \widetilde{\varphi''_{sc}}(a_1\otimes_R b_1) + \cdots + \widetilde{\varphi''_{sc}}(a_n\otimes_R b_n) \\&= \widetilde{\varphi''_{sc}}(t). \end{align*}
Additionally note that $\varphi'$ is additive in both slots, so we get a well-defined map with $\varphi(t\otimes_S c) := \varphi'(t, c)$ satisfying $\varphi((a\otimes_R b) \otimes_S c) = a \otimes_R (b \otimes_S c)$.
You can repeat all of this to similarly define an inverse mapping $\psi\colon A \otimes_R (B \otimes_S C)\to (A\otimes_R B) \otimes_S C$ satisfying $\psi(a \otimes_R (b \otimes_S c)) = (a\otimes_R b) \otimes_S c$.
Any tensor is a sum of pure tensors, so these mappings are indeed inverses.