By Lemma 10.86.1, we get that $0\to \lim K/(I^nR^t\cap K)\to (R^t)^\wedge \to M^\wedge \to 0$ is exact. By Artin-Rees lemma, $\lim K/(I^nR^t\cap K)=K^\wedge$ . Therefore we get $0\to K^\wedge\to (R^t)^\wedge \to M^\wedge \to 0$ is exact as required.

Consider a presentation of module M (which is f.g. by assumption) $0\to K\to R^t\to M\to 0$ We get that $0\to K/(I^nR^t\cap K)\to R^t/(I^nR^t)\to M/(I^nM)\to 0$ is exact. (Here $K$ is viewed as a submodule of $R^t$.) One sees this by observing that $I^nR^t$ maps surjectively onto $I^nM$.

By Lemma 10.86.1, we get that $0\to \lim K/(I^nR^t\cap K)\to R^\wedge \to M^\wedge \to 0$ is exact. By Artin-Rees lemma, $\lim K/(I^nR^t\cap K)=K^\wedge$ Therefore we get $0\to K^\wedge\to R^\wedge \to M^\wedge \to 0$ is exact as required.

A correction in the proof of Lemma 10.96.2:

Here I think you mean an arbitrary ideal $J$ in R (and not specifically the ideal $I$ w.r.t. which $R$ is completed): $J \otimes _ R R^\wedge \to R \otimes _ R R^\wedge = R^\wedge$.