We define two modules $M,M'\in Mod-R$ projectively equivalent if there are projective modules $P,P'$ s.t. $M\oplus P\cong M'\oplus P'$. The problem is the following: suppose to have two exact sequences: $$0\to K\to P_{n-1}\to...\to P_0\to M\to 0$$ and $$0\to K'\to P_{m-1}'\to...\to P_0'\to M\to0$$ such that $P_i,P_j'$ are projective. Then $K$ and $K'$ are projectively equivalent.
This fact is true for $n,m=1$ by Schanuel lemma: if there are two exact sequences: $$0\to K\to P\to M\to 0$$ $$0\to K'\to P'\to M\to 0$$ With $P,P'$ projective then $K\oplus P'\cong K'\oplus P$ so are equivalent. So the idea is to use induction (probably on $n+m$) by building two sequences shorter and apply the inductive hypothesis.
As Jeremy Rickard explain in a comment, this is not necessarily true when $m \neq n$. Here is a very simple example for $n = -1$ and $m = 0$ (finding example with nonnegative integers involves finding modules with positive projective dimension, so you can't work over a PID, but the idea is the same): $$0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$$ $$0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$$ Since $\mathbb{Z}/2\mathbb{Z}$ is not projective but $\mathbb{Z}$ is, they are not projectively equivalent.
When $m=n$ however you can use induction. You already have the case $n=1$, let's assume the proposition proven for a given $n \ge 1$ and take two exact sequences: $$0\to K \to P_n \to \dots \to P_0' \to M\to 0$$ $$0 \to K' \to P_n' \to \dots \to P_0' \to M\to 0$$
Let $$L = \operatorname{ker}(P_{n-1} \to P_{n-2}) = \operatorname{im}(P_n \to P_{n-1})$$ and define similarly $L'$ with primes everywhere (assuming $P_{-1} = P_{-1}' = M$ in case $n = 1$). You then have two exact sequences: $$0 \to L \to P_{n-1} \to \dots \to P_0 \to M \to 0$$ and the same with primes everywhere. By the induction hypothesis, it follows there exists some projective modules $P$ and $P'$ such that $L \oplus P \cong L' \oplus P'$.
By definition of $L$ you have a short exact sequence $$0 \to K \to P_n \to L \to 0$$ and the same with primes everywhere. Add $P$ and $P'$ to these to get two short exact sequences: $$0 \to K \to P_n \oplus P \to L \oplus P \to 0$$ $$0 \to K' \to P_n' \oplus P' \to L' \oplus P' \to 0$$ but since $L \oplus P \cong L' \oplus P'$ and both $P_n \oplus P$ and $P_n' \oplus P'$ are projective, by Schanuel's lemma $K$ and $K'$ are projectively equivalent. In fact you can prove more precisely that: $$K \oplus P_0 \oplus \dots \oplus P_n \cong K' \oplus P_0' \oplus \dots \oplus P_n'.$$