I am looking at Exercise I.2.3 in Weibel's $K$-book. Here is the statement:
If $I$ is an ideal of a ring $R$, we form the augmented ring $R\oplus I$ and let $K_0(R,I)$ denote the kernel of $K_0(R\oplus I)\to K_0(R)$. Show there is an exact sequence $$\operatorname{GL}(R)\to\operatorname{GL}(R/I)\overset\partial\to K_0(R,I)\to K_0(R)\to K_0(R/I).$$
I'm having trouble telling what these maps are meant to be. I don't even see what a natural map $K_0(R,I)\to K_0(R)$ would be, except for the restriction of $K_0(R\oplus I)\to K_0(R)$ (but then we would get the zero map, since $K_0(R,I)$ is the kernel of this map by definition; and then this result would imply that $K_0(R)\to K_0(R/I)$ is always injective, which seems ridiculous). I also don't see what $\partial$ should be; I guess maybe there should be some natural Milnor square $$\require{AMScd}\begin{CD} R\oplus I @>>> R \\@VVV & @VVV \\ ? @>>> R/I \end{CD}$$ along which we could patch free modules and get a resulting map $\operatorname{GL}_n(R/I)\to K_0(R\oplus I)$ for each $n$, and try to glue these. But I don't see what this would be (I'm even shaky on what the multiplication is on $R\oplus I$; the only reasonable option seems to be $(r,a)\,(r',a') := (rr',r'a+ra')$ but this isn't stated explicitly in the text).
Personnally, I have another definition of $K_0(R,I)$ (it probably coincides with your definition, though, see the end of this answer...). Let $D=R\times_I R=\{ (x,y)\in R\times R \mid \bar{x}=\bar{y}\in R/I\}$
Then we have a fiber square of rings $$\require{AMScd}\begin{CD} D @>>> R \\@VVV & @VVV \\ R @>>> R/I \end{CD}$$ where the top map is the second projection, and the first vertical map is the first projection. Then $K_0(R,I)$ is the kernel of the induced map $K_0(\pi_2):K_0(D)\to K_0(R)$ by definition.
Now, you may restricts $K_0(\pi_1)$ to get a map $K_0(R,I)\to K_0(R).$
If $C\in GL_n(R/I)$, recall that $(R^n,R^n,C)=\{(x,y)\in R^n\times R^n\mid C\bar{x}=\bar{y}\}$. You now may define $\partial(C)=[(R^n,R^n,C)]-[(R^n,R^n,I_n)]\in K_0(D)$, and show that it lies in fact in $K_0(R,I)$.
How to connect the dots with Weibel's definition.
Certainly, you have a bijection between $R\oplus I$ and $D$ given by $(r,a)\mapsto (r,r+a)$. It obviously preserves sums. If you want this bijection to preserve products, the only way to proceed is to set $(r,a)(r',a')=(rr',ar'+a'r+aa')$. The unit element will be $(1_R,0_R)$, and the bijection will become an isomorphism of rings with $1$.