Showing that a problem in $NP$ admits two distinct verifiers that satisfy additional constraints
Since $S \in \mathsf{NP}$, there is a non-deterministic poly-time Turing machine $T^*$ that decides $S$.
Given $x \in S$, let $y^* = \langle y^*_1, y^*_2, \dots\rangle \in \{0,1\}^*$ be the list of non-deterministic choices of $T^*$ that lead to an accepting state in the execution of $T^*(x)$.
It is easy to see that $|y^*| = O(\mathrm{poly}(x))$ and that $y^*$ is a certificate for the the polynomial-time (deterministic) verifier $V^*$ that, with input $V^*(x,y^*)$, checks that the (unique) computation path of $T^*(x)$  when the non-deterministic choices are those in $y^*$ leads to an accepting state.
Define $y_1 =\langle 0, y^*_1, y^*_2, \dots \rangle$ and  $y_2 =\langle 1, y^*_1, y^*_2, \dots \rangle$.
When $y=\langle y_0, y_1, y_2, \dots \rangle$, the verifier $V_i(x, y)$  behaves as follows: if $y_0 \neq i-1$, it rejects. Otherwise $V_i(x, y)$ simulates $V^*(x, \langle y_1, y_2, \dots\rangle)$ and accepts iff $V^*$ accepts. It is clear that $V_1$ and $V_2$ are polynomial-time verifiers for $S$ for the certificates $y_1$ and $y_2$, respectively.
It remains to show that they satisfy the additional constraints from your question.
Notice that, if $V_1(x,y)$ accepts, then $y_0=0$, and hence $V_2(x,y)$ rejects since $y_0 = 0 \neq 1 = 2-1$.
Similarly, if $V_2(x,y)$ accepts, then $y_0=1$, and hence $V_1(x,y)$ rejects since $y_0 = 1 \neq 0 = 1-1$.