I think the proof is (a little) faulty...
As a counter-example to the argument (but not to the result!), take $$A= k[f_1, f_2, x_1,x_2],$$ subject to the constraints
$$f_1+ f_2 =1 \text{ and } f_1f_2( x_1f_2 - x_2f_1) = 0.$$
Then, $U_1 \cup U_2 = X$, where $U_i = \mathop{\text{spec}} A_{f_i}$, as $f_1 + f_2 =1$.
By construction, if $s_i =x_i/f_i$, $s_1/1 = s_2/1$ in $A_{f_1f_2}$.
Following the notation of the proof, we can take $$a_i=1\in A,$$ and likewise for the exponents: $$m_1=m_2=n_{12}=1.$$
The argument suggests that one set $s= x_1 + x_2$,
and argues (it looks erroneously) that $s/1 = s_1 \in A_{f_1}$:
$$\begin{align*}
s/1 &= x_1/1 + x_2/1\\
&= s_1f_1 + x_2/1 \\
&= s_1 + (x_2/1 - f_2s_1)\\
&= s_1 + 1/f_1 \Big(x_2f_1 - f_2x_1\Big).\\
\end{align*} $$
So, as you say, for $s/1= s_1 \in A_{f_1}$, that means that
$$ f_1^n (x_2f_1 -f_2x_1) = 0 \in A,
\text{ for some }n,$$
which is not implied by the defining relations.
But! In this example,
$$ f_1f_2(x_1f_2-x_2f_1 ) =(x_1f_1) f_2^2 - (x_2f_2)f_1^2 = 0 \in A.$$
Furthermore,
$$s_i= x_i/f_i= x_if_i/ f_i^2\in A_{f_i},$$
and there are $a_1,a_2\in A$, such that $a_1f_1^2+ a_2f_2^2=1$, and proceeding with $s = a_1 (x_1f_1) + a_2 (x_2f_2)$ does work...
Now, in general:
First, to hide exponents, write the $s_i= x_i/f^{m_i}_i$ as $ x_i/g_i$, with $g_i = f_i^{m_i}$.
Then - bearing in mind that $A_{f_i}= A_{g_i}$ - the $U_i= \mathop{\text{spec}}A_{g_i}$ constitute the same cover of $X$, and we can continue with $g_i$ in place of $f_i$.
By quasi-compactness, (i.e., only consider the $i$ and $j$ from a finite cover), there exists a fixed $n$, sufficiently large, such that
$$ (g_ig_j)^n(x_ig_j - x_jg_i)= (g_i^nx_i)g_j^{n+1}- (g_j^nx_j)g_i^{n+1} = 0.$$
Relabel again to ease notation/algebra: write $s_j = y_j/h_j$, with $y_j=x_jg_j^n$, and $h_j=g_j^{n+1}$.
Then the previous displayed equation gives $$ y_ih_j-y_jh_i = 0 \in A.$$
With $a_j \in A$ chosen so that $ a_1h_1+ \cdots + a_kh_k = 1$, take
$$ s = a_1y_1 + \cdots a_ky_k \in A.$$
The steps of the argument (as in the original question) show that
$$s/1 = s_i \in A_{f_i}=A_{h_i},$$
as desired.
I think is useful to write correctly our discussion in comments. Let me consider a slightly more general statement, which might be useful.
Let $X=\operatorname{Spec}(A)$, and $U_1=X_{f_1}$, $U_2=X_{f_2}$, $V=X_g$, such that $V \subseteq U_1 \cap U_2$. If $s_i=x_i/f_i^{m_i} \in A(U_i)=A_{f_i}$ agree on $V$, then
$$ g^k(x_1f_2^{m_2}-x_2f_1^{m_1})=0 $$
for some $k >0$.
The intuition here is that this is the condition for the statement $x_1/f_1^{m_1}=x_2/f_2^{m_2}$ in $A_g$. But it is not obvious how to interpret "$1/f_1$ in $A_g$". One needs to use the following remark.
If $X_g \subseteq X_f$, then $g^s=uf$ for some $s>0$ and some $u \in A$.
This way, we can write $g^{d_1}=u_1f_1$, $g^{d_2}=u_2f_2$. Therefore, the map $A_{f_1}=A(U_1) \to A(V)=A_g$ maps $x_1/f_1^{m_1}$ to $x_1u_1^{m_1}/g^{d_1m_1}$, and similarly with $x_2/f_2^{m_2}$.
Now, by assumption we have that the images of $s_1$ and $s_2$ agree on $A_g$. This is, we have
$$ \frac{x_1u_1^{m_1}}{g^{d_1m_1}} = \frac{x_2u_2^{m_2}}{g^{d_2m_2}} $$
in $A_g$. This is equivalent to
$$ 0=g^r(x_1u_1^{m_1}g^{d_2m_2} - x_2 u_2^{m_2}g^{d_1m_1}) $$
in $A$, for some $r>0$. Now, using that $g^{d_1}=u_1f_1$ and $g^{d_2}=u_2f_2$, we can rewrite
\begin{align*}
0 =& g^r(x_1u_1^{m_1}g^{d_2m_2} - x_2 u_2^{m_2}g^{d_1m_1}) \\
=& g^r(x_1u_1^{m_1}u_2^{m_2}f_2^{m_2} - x_2 u_2^{m_2}u_1^{m_1}f_1^{m_1}) \\
=& g^r u_1^{m_1}u_2^{m_2}(x_1f_2^{m_2}-x_2f_1^{m_1}).
\end{align*}
Multiply everything by $f_1^{m_1}f_2^{m_2}$ to get
$$ 0 = g^{r+d_1m_1+d_2m_2}(x_1f_2^{m_2}-x_2f_1^{m_1}). $$
This is exactly the original condition, taking $k=r+d_1m_1+d_2m_2$.
(As a remark, your original question is this one with $g=f_1f_2$, $d_1=d_2=1$, $u_1=f_2$, $u_2=f_1$. This is what I sketched in the comments.)
Best Answer
It means either of the following things, which are all canonically isomorphic (here, canonically = uniquely if you require that it respects the map coming from $M$):
$M[(f_if_j)^{-1}]$, $M[f_i^{-1}][f_j^{-1}]$, $M[\{f_i,f_j\}^{-1}]$
In particular, you have a morphism $M[f_i^{-1}]\to M[f_i^{-1}][f_j^{-1}]$ and in particular $m_i$ (an $m_j$) is sent to someone in the latter.
If you want to view it as $M[(f_if_j)^{-1}]$, then indeed, you will have that the image of $m_i = \frac{n_i}{f_i^{k_i}}$ is $\frac{n_if_j^{k_i}}{(f_if_j)^{k_i}}$
But you donw't have to view it that way, you can also simply say it's $\frac{m_i}{1}$ for instance (that's the $M[f_i^{-1}][f_j^{-1}]$ point of view) or just $\frac{n_i}{f_i^{k_i}}$ (that's the $M[\{f_i,f_j\}^{-1}]$ point of view)