If $L_1,L_2\in \mathrm{NP}$ and $w\in L_1$ or $w\in L_2$ then can $L_1\cup L_2=L$'s verifier use the same certificate $c$ for $w$? I read the following solution for Showing that $\mathrm{NP}$ is closed under union
and they used the same $c$ for both the verifies $V_1$ and $V_2$. Why is it correct?

Let $L_1$ and $L_2$ be languages in $\mathrm{NP}$. Also, for $i = 1, 2$ let $V_i(x, c)$ be
an algorithm that, for a string $x$ and a possible certificate $c$,
verifies whether $c$ is actually a certificate for $x \in L_i$. Thus, $V_i(x,c) = 1$ if certificate $c$ verifies $x \in L_i$, and $V_i(x, c) = 0$ otherwise.
Since both $L_1$ and $L_2$ are both in $\mathrm{NP}$, we know that $V_i(x, c)$ terminates
in polynomial time $O(|x|^d)$ for some constant $d$. To show that $L_3 = L_1 \cup L_2$ is also in $NP$, we will construct a polynomial-time verifier $V_3$ for $L_3$. Since a certificate $c$ for $L_3$ will have the property that
either $V_1(x, c) = 1$ or $V_2(x, c) = 1$, we can easily construct a
verifier $V_3(x, c) = V_1(x, c) \lor V_2(x, c)$. Clearly then $x \in L_3$ if and
only if there is a certificate $c$ such that $V_3(x, c) = 1$. Notice also
that the new verifier $V_3$ will run in time $O(2(|x|^d))$, which is
polynomial. Therefore, the union $L_3$ of two languages in $\mathrm{NP}$ is also in $\mathrm{NP}$, so $\mathrm{NP}$ is closed under union.

taken from here.
$M_1,M_2$ TM which accept $w$ can accept $w$ for different reasons so we can't claim that $c_1=c_2$
Questions:

Is it legal to use the same certificate for both $V_1,V_2$ in the answer? Why?

Is a verifier by definition is deterministic TM?

In the above answer, does $V_3$ runs $x,c$ on both $V_1$ and $V_2$ in the worst case?
Assume that $c_1$ and and $c_2$ are polynomial size certificates for $x \in V_1$ and $x \in V_2$. Then define $c=c_1\#c_2$, a new certificate formed by concatenation of $c_1$ and $c_2$ which is clearly polynomial size. Then $V_1$ and $V_2$ still can use $c$ as a certificate to verify $x\in V_1$ and $x\in V_2$. $V_1$ will use the left part of $c$ and the $V_2$ will use the right part of $c$. 


  Is a verifier by definition is deterministic TM?


Yes, the verifier is a deterministic TM, by definition.


  In the above answer, does $V_3$ runs $x,c$ on both $V_1$ and $V_2$ in the worst case?


You can treat $V_3$ as a TM which invokes (as subroutines) $V_1(x,c)$ and $V_2(x,c)$, so if $V_1$'s worst case is $O(f)$ and $V_2$'s worst case is $O(g)$ then $V_3$'s worst case is either $O(f)$ or $O(g)$ which is polynomial.