Proof of Quillen's patching theorem The following are from Lam's book Serre's problem on projective modules, on page 163 and 164.
For any ring $A$, the notation $m \in \mathscr{R}^A(A[t_1, \dots, t_n])$ means that there exists a $A-$
module $N$ such that $M \cong A[t_1, \dots, t_n] \otimes_A N$.
Theorem 1.6. (Quillen’s Patching Theorem) Let $R$ be a commutative ring. Let $A$
be any (not necessarily commutative) $R$-algebra, and let $M$ be a finitely presented
$A[t_1, \dots, t_n]$-module. Then:
$(A_n)$ Q(M) := $\{g \in R : M_g \in \mathscr{R}^{A_g}(A_g[t_1, \dots, t_n])\}$ is an ideal in $R$ (called the Quillen ideal of $M$).
$(B_n)$ If $M_m \in \mathscr{R}^{A_m}(A_m[t_1, \dots, t_n])$ for every maximal ideal $m \in$ Max $R$, then $M \in \mathscr{R}^{A}{A[t_1, \dots, t_n]}$.


In step 3, what does "$u_i$ reduces modulo $t$ to the identify map of $N_{f_i}$" mean?
Why "After composing this with a suitable automorphism on $N_{f_i[t]}$, we
may assume that $u_i$ reduces modulo t to the identify map of $N_{f_i}$"?
Why $θ$ reduces to the identity mod $t$?
 A: It seems that there is a typo and what is meant is "$u_i$ reduces modulo $t$ to the identity map of $N_{f_i}$".
The idea is that $u_i$ induces an isomorphism modulo $t$, $\overline{u_i}: M_{f_i}/tM_{f_i} \rightarrow N_{f_i}=(M/tM)_{f_i}$, and there is a canonical identification $ M_{f_i}/tM_{f_i} \simeq (M/tM)_{f_i},$ so we may treat $\overline{u_i}$ as an automorphism $\alpha:=\overline{u_i}: N_{f_i} \rightarrow N_{f_i}$. The idea is to post-compose $u_i$ with the automorphism $$\alpha^{-1}[t]: N_{f_i}[t] \rightarrow N_{f_i}[t],$$
the extension of $\alpha^{-1}$ determined by the rule $t \mapsto t$.
As for the second question, note that since now $u_i$ reduce to the identity mod $t$ in the above sense, the same is true for their further localizations ${u_0}_{f_1}$ and ${u_1}_{f_0}$ (as in the above paragraph, one might need to use a canonical isomorphism expressing that localization and quotient commute to make sense of the statement). Since $\theta$ is defined as their composite $\theta={u_1}_{f_0} \circ ({u_0}_{f_1}^{-1})$, the same is true for $\theta$ (modulo $t$, it is just the map $\mathrm{id}_{N_{f_0f_1}} \circ \mathrm{id}_{N_{f_0f_1}}^{-1}=\mathrm{id}_{N_{f_0f_1}}$).
