Let M be an R-module, where R is a commutative ring with unity. Take two elements $s,t \in R$ such that $Rs+Rt=R$. Then we want to show below commutative diagram is fiber product diagram:
Basically we need to show for any $m_i/s^i \in M_s$ and $m_j/t^j \in M_t$ mapping to same element in $M_{st}$, there exists a unique element in $M$ which is pre-image of these two elements.
For $m_i/s^i$ and $m_j/t^j$ mapping to same element in $M_{st}$ we get $ s^k t^{k^{'}} [t^jm_i-s^im_j]=0$, for some $k,k^{'}\in \mathbb{N} $.
Now for any $n_1,n_2 \in \mathbb{N}$, using $Rs+Rt=R$ we get $Rs^{n_1}+Rt^{n_2}=R$, else $Rs^{n_1}+Rt^{n_2} \subset M$ for some maximal ideal $M$, then $s^{n_1}, t^{n_2} \in M \implies s,t \in M \implies M = R$ which is contradiction. So we have $a_{n_1}s^{n_1}+b_{n_2}t^{n_2}=1$ for some $a_{n_1},b_{n_2}\in R$.
Using both the facts we obtain: $s^k[s^i x_i -m_i]=0$ and $t^{k^{'}}[t^j y_j -m_j]=0$. Where $x_i = \{ a_im_i+b_{k^{'}+j}t^{k^{'}}m_j \}$, $y_j = \{ a_{k+i}s^{k}m_i + b_jm_j\}$ are pre-images of $m_i/s^i$ and $m_j/t^j$ respectively.
So whenever images of $m_i/s^i$ and $m_j/t^j$ are equal, we get $m_i/s^i = x/1$ and $m_j/t^j=y/1$ for some $x=x_i, y=y_j \in M$.
Now, $x/1,y/1$ have same images implies $s^kt^{k^{'}}[x-y]=0$. Similar as above we manipulate to get: $s^k[m-x]=0$ and $t^{k^{'}}[m-y]=0$ where $m=\{a_ks^kx+b_{k^{'}}t^{k^{'}}y \}$. So $m\in M$ is a pre-image of $x/1$ and $y/1$.
Now how to prove that such $m$ is unique.
Here's how you do this, lets let $t=t_1$ and $s=t_2$. Then suppose that $m_1/t_1^{k_1}\in M_{t_1}$ and $m_2/t_2^{k_2}\in M_{t_2}$ such that: \begin{align} \frac{m_1\cdot t_2^{k_1}}{(t_1t_2)^{k_1}}=\frac{m_2\cdot t_1^{k_2}}{(t_1t_2)^{k_2}} \end{align} This implies that there is $K$ such that: $$(t_1t_2)^K\left[(t_1t_2)^{k_2}\cdot m_1\cdot t_2^{k_1}- (t_1t_2)^{k_1}\cdot m_2\cdot t_1^{k_2}\right]=0$$ Multiply this expression by $t_1^{k_1}\cdot t_2^{k_2}$ to obtain that: $$(t_1t_2)^K\left[m_1t_2^{k_2}-m_2t_1^{k_1}\right]=0$$ for some new $K$. We set: $$h_i=m_it_i^K$$ and note that since $\langle t_1\rangle +\langle t_2\rangle=R$, there exist $c_i$ such that $c_1t_1^{K+k_1}+c_2t_2^{K+k_2}=1$. We thus define $b\in M$ by: $$b=c_1h_1+c_2h_2$$ Now we claim that: $$\frac{c_1h_1+c_2h_2}{1}=\frac{m_1}{t_1^{k_1}}$$ So we need to find an $M$ such that: $$t^M_1\left[t_1^{k_1}\cdot (c_1h_1+c_2h_2)-m_1\right]=0$$ Well, set $M=K$ and note that: \begin{align}T_1^Kt_1^{k_1}\cdot (c_1h_1+c_2h_2)=&t_1^{k_1}c_1m_1t_1^Kt_1^K+t_1^{k_1}c_2m_2t_1^Kt_2^K \\ =&t_1^{k_1}c_1m_1t_1^Kt_1^K+c_2m_1t_2^{k_2}t_1^Kt_2^K\\ =&m_1t_1^K\left(c_1t_1^{K+k_1}+c_2t_2^{K+k_2}\right)\\ =&m_1t_1^K\end{align} so $b/1=m_1/t_1^{k_1}$, and similarly for $m_2/t_2^{k_2}$.
Now suppose that $\theta\in M$ also satisfies $\theta/1=m_1/t_1^{k_1}, m_2/t_2^{k_2}$, then $(\theta-b)/1=0$ in both situations. Let $f=\theta-b$, then in both situations $f/1=0$, so there exists some $l_i$ such that $t_i^{l_i}\cdot f=0$. Again since $\langle s\rangle +\langle t\rangle =1$, there exist $c_i$ such that $c_1t_1^{l_1}+c_2t_2^{l_2}=1$, so: $$f=c_1\cdot t_1^{l_1}f+c_2\cdot t_2^{l_2}f=0+0=0$$ so any other element that restricts to $m_i/t_i^{k_i}$ for each $i$ must be equal to $b$.
I have no idea why but this is strikingly similar to proving that the sheaf on $\operatorname{Spec} A$ defined by $D(f)\mapsto A_f$ for some commutative ring $A$ is a sheaf on a base.