Yes, this is true.

Let $K_0(R)$ be the Grothendieck group of $R$ and $\widetilde K_0(R)=K_0(R)/\langle [R]\rangle$. Then $[P]=0$ in $\widetilde K_0(R)$ if and only if $P$ is stably free.

For any $FP$ $R$-module $M$ having the following resolution consisting of finitely generated projective $R$-modules $$0\to P_k\to \ldots \to P_1\to P_0\to M\to 0,$$ we denote $$\chi_u^R(M)=\sum_{i=0}^k (-1)^i [P_i]\in \widetilde K_0(R).$$ Observe that $\chi_u^R(M)=0$ if and only if $M$ is $FL$.

Thus, in order to prove the claim it is enough to show that if $0\to M_1\to M_2\to M_3\to 0$ is an exact sequence of $FP$ $R$-modules, then $\chi_u(M_2)=\chi_u(M_1)+\chi_u(M_3)$.

This can easily be done by induction on $pd_R(M_1)+pd_R(M_2)+ pd_R(M_3)$, where $pd_R(M)$ is the projective dimension of $M$.

Yes, this is true. See Bourbaki, N. Éléments de mathématique. Algèbre. Chapitre 10. Algèbre homologique. (French) [Elements of mathematics. Algebra. Chapter 10. Homological algebra], Theorem 3.9.1

Let $K_0(R)$ be the Grothendieck group of $R$ and $\widetilde K_0(R)=K_0(R)/\langle [R]\rangle$. Then $[P]=0$ in $\widetilde K_0(R)$ if and only if $P$ is stably free.

For any $FP$ $R$-module $M$ having the following resolution consisting of finitely generated projective $R$-modules $$0\to P_k\to \ldots \to P_1\to P_0\to M\to 0,$$ we denote $$\chi_u^R(M)=\sum_{i=0}^k (-1)^i [P_i]\in \widetilde K_0(R).$$ Observe that $\chi_u^R(M)=0$ if and only if $M$ is $FL$.

Thus, in order to prove the claim it is enough to show that if $0\to M_1\to M_2\to M_3\to 0$ is an exact sequence of $FP$ $R$-modules, then $\chi_u(M_2)=\chi_u(M_1)+\chi_u(M_3)$.

This can easily be done by induction on $pd_R(M_1)+pd_R(M_2)+ pd_R(M_3)$, where $pd_R(M)$ is the projective dimension of $M$.

Yes, this is true.

Let $K_0(R)$ be the Grothendieck group of $R$ and $\widetilde K_0(R)=K_0(R)/\langle [R]\rangle$. Then $[P]=0$ in $\widetilde K_0(R)$ if and only if $P$ is stably projective.

For any $FP$ $R$-module $M$ having the following resolution consisting of finitely generated projective $R$-modules $$0\to P_k\to \ldots \to P_1\to P_0\to M\to 0,$$ we denote $$\chi_u^R(M)=\sum_{i=0}^k (-1)^i [P_i]\in \widetilde K_0(R).$$ Observe that $\chi_u^R(M)=0$ if and only if $M$ is $FL$.

Thus, in order to prove the claim it is enough to show that if $0\to M_1\to M_2\to M_3\to 0$ is an exact sequence of $FP$ $R$-modules, then $\chi_u(M_2)=\chi_u(M_1)+\chi_u(M_3)$.

This can easily be done by induction on $pd_R(M_1)+pd_R(M_2)+ pd_R(M_3)$, where $pd_R(M)$ is the projective dimension of $M$.

Yes, this is true.

Let $K_0(R)$ be the Grothendieck group of $R$ and $\widetilde K_0(R)=K_0(R)/\langle [R]\rangle$. Then $[P]=0$ in $\widetilde K_0(R)$ if and only if $P$ is stably free.

For any $FP$ $R$-module $M$ having the following resolution consisting of finitely generated projective $R$-modules $$0\to P_k\to \ldots \to P_1\to P_0\to M\to 0,$$ we denote $$\chi_u^R(M)=\sum_{i=0}^k (-1)^i [P_i]\in \widetilde K_0(R).$$ Observe that $\chi_u^R(M)=0$ if and only if $M$ is $FL$.

Thus, in order to prove the claim it is enough to show that if $0\to M_1\to M_2\to M_3\to 0$ is an exact sequence of $FP$ $R$-modules, then $\chi_u(M_2)=\chi_u(M_1)+\chi_u(M_3)$.

This can easily be done by induction on $pd_R(M_1)+pd_R(M_2)+ pd_R(M_3)$, where $pd_R(M)$ is the projective dimension of $M$.