Let $R$ be a ring with unit and $I$ a two sided ideal. We define the double ring $D(R,I)$ as the kernel pair for $q : R \to R/I$, i.e. as
$$ D(R,I) = \{(x,y) \in R^n \times R^n : x = y \pmod{I} \} $$
Likewise, if $A \in M_n(R)$ is such that its reduction $\dot{A} \in M_n(R/I)$ is invertible, then we define $$ R^n \times_{\dot{A}} R^n = \{(x,y) \in R^n \times R^n : xA = y \pmod{I} \} $$ as the pullback of $R^n \xrightarrow{q \times \cdots \times q} R/I$ and $R^n \xrightarrow{ - \cdot A} R^n \xrightarrow{q \times \cdots \times q} R/I$.
The latter are $D(R,I)$ modules, via $(r,r')(x,y) = (rx,r'y)$. Moreover these are always projective: the block diagonal matrix $\dot{A} \oplus (\dot{A})^{-1}$ is always a product of elementary $R/I$ matrices, and so it can be lifted to $B \in GL(R)$. Taking $C$ such that $\dot{C} = (\dot{A})^{-1}$ gives
$$ R^n \times_{\dot{A}} R^n \oplus R^n \times_{\dot{C}} R^n \simeq R^n \times_{\dot{A} \oplus \dot{A}^{-1}} R^n \simeq R^n \times_{\dot{B}} R^n \xrightarrow{(x,y) \mapsto (xB,y)} D(R,I)^n $$
On the other hand, any $D(R,I)$-module can be made an $R$ module in two ways, by tensoring with $R$ thought of as a $D(R,I)$-module via pullback of the canonical projections $p_1,p_2$.
What I want to prove is the following,
If $P$ is a projective $D(R,I)$-module such that $P \otimes_{D(R,I)} p_1^\ast R \simeq P \otimes_{D(R,I)} p_2^\ast R \simeq R^n$ as $R$-modules, there exists $A \in M(n,R)$ such that $P \simeq R^n \times_{\dot{A}} R^n$.
which is claimed to be clear in the proof I am reading, but I am having a hard time proving it nonetheless.
This appears in Theorem 2.5.4 of Rosenberg's Algebraic K-theory and its applications, namely, the result exhibits the exact sequence which connects $K_1$ with $K_0$.