Definition of localization of a ring is slightly redundant When $U^{-1}R$ is defined, we define $(r_1, u_1) \sim (r_2,u_2)$ if there exists $u \in U$ such that $u(r_1 u_2 - r_2 u_1) = 0.$
First observe that $U$ can't have $0$ divisors, for if $tu = 0,$ then $1/u$ should be well-defined, yet it's $t/(tu) = t/0,$ which doesn't make sense (*). Thus, we can remove $u$ from the definition of $\sim$ and write $u_1 r_2 - u_2 r_1 = 0.$
My project leader couldn't find anything wrong with the argument, but given that thousands of people have looked over the definition throughout history, they must be right and I must be wrong. The only place there's room for a mistake is in (*). Let's continue to suppose $1/u$ is well-defined. Then $1/u = t/0 = tu/0 = 0/0.$ We have $0/0 + 0/0 = 0/0 \Rightarrow 0/0 = 0.$ Finally, $1 = u/u = u(1/u) = 0u = 0.$ Now where's the mistake?
 A: The definition is not superfluous; you do want that extra factor in general.
If $0\in U$, then $U^{-1}R$ will be the zero ring. There is no error there: any two pairs $(r,u)$ and $(r',u')$ will be equivalent, so $U^{-1}R$ will have one element: it is the zero ring, where $0=1$.
Here is an example where you have zero divisors in $U$: let $R=\mathbb{Z}/6\mathbb{Z}$, and let $U=\{2,4\}$. (I will omit the bars to denote the equivalence classes for simplicity)
Note that both elements of $U$ are zero divisors. The ring $U^{-1}R$ is isomorphic to $\mathbb{F}_3$, the field of three elements. The images of $2$ and $4$ are both invertible in $U^{-1}R$.
Indeed, the elements of $U^{-1}R$ are equivalence classes of pairs $(r,u)$ with $r\in R$ and $u\in U$. There are twelve such pairs.
Let's take $(0,2)$; we have $(0,2)\sim(r,u)$ if and only if there exists $u''\in U$ such that $u''(0u-2r)\equiv 0\pmod{6}$, if and only if $2u''r\equiv 0\pmod{6}$, if and only if $u''r\equiv 0\pmod{3}$, if and only if $r\equiv 0\pmod{3}$ (since $u''\in U$ is invertible modulo $3$). So the class of $(0,2)$ is
$$[0,2] = \{(0,2), (0,4), (3,2), (3,4)\}.$$
Now, $(1,2)\sim (r,u)$ if and only if $u''(u-2r)\equiv 0\pmod{6}$ for $u''\in U$. Since $u''$ is even, we get $u-2r\equiv 0\pmod{3}$, or $u+r\equiv 0\pmod{3}$. This gives
$$[1,2] = \{(1,2), (4,2), (2,4), (5,4)\}.$$
Next $(2,2)\sim (r,u)$ if and only if $u''(2u-2r)\equiv 0\pmod{6}$; this gives $u''(u-r)\equiv 0\pmod{3}$, and since $u''\in U$ is invertible modulo $3$, this gives $u\equiv r\pmod{3}$. Thus,
$$[2,2] = \{(2,2), (5,2), (1,4), (4,4)\}.$$
This gives all the elements. So $U^{-1}R$ consists of the "fractions"
$$\begin{align*}
\frac{0}{2} &= \frac{0}{4} = \frac{3}{2} = \frac{3}{4},\\
\frac{1}{2} &= \frac{4}{2} = \frac{2}{4} = \frac{5}{4},\\
\frac{2}{2} &= \frac{5}{2} = \frac{1}{4} = \frac{4}{4}.
\end{align*}$$
$\frac{0}{2}$, $\frac{1}{2}$, and $\frac{2}{2}$.  Note that even though $2$ is a zero divisor, the image of $2$ under the canonical map $\varphi\colon R\to U^{-1}R$ which sends $r$ to $\frac{2r}{2} = \frac{4r}{4}$ gives $\frac{4}{2}$, which is of course the inverse of $2$.
The error in your argument is that although $2$ is a zero divisor in $R$, its "complementary divisor" which you call $t$ is $3$. But $3$ is not in $U$, so that you cannot form the fraction $\frac{1}{st}$. Only elements of $U$ are allowed in the "denominator".

Now, in this example it turns out that the extra factor in the definition does not matter. But in general, if you do not have that extra factor, then the definition may fail to yield an equivalence relation. The problem lies in transitivity of the relation.
Suppose $(r,u)\sim (s,v)$, and $(s,v)\sim(t,w)$, under the definition that omits $u''$. Then we have $rv-us=0$ and $sw-vt=0$. We want to show that $rw-ut=0$; the usual argument is to multiply the first equality by $w$ and the second by $u$. So we have $rvw-usw=0$ and $suw-vtu=0$. Adding them gives $rvw-vtu = v(rw-ut)=0$. If $v$ is not a zero divisor, then this implies that $rw-ur=0$ and we are done. But if $v$ is a zero divisor (which we have seen above can happen without a contradiction), then this could be a problem. So let us construct an example where this happens by tweaking the example above.
Take $R=\mathbb{Z}/6\mathbb{Z}$, but now let $U=\{1,2,4\}$. This will yield the same ring $U^{-1}R$, since $1$ is already invertible.
Now we have $(3,1)\sim(0,2)$, since $(3)(2) - (1)(0)=0$ in $R$. Likewise, $(0,2)\sim(0,1)$, since $(0)(1)-(0)(2)=0\in R$. But is $(3,1)\sim(0,1)$? If we don't have that extra factor, then it is not: because $(3)(1)-(1)(0)\neq 0$. So we would have $(3,1)\sim(0,2)$, and $(0,2)\sim(0,1)$, but $(3,1)\not\sim (0,1)$. We don't have a transitivie relation,, and hence not an equivalence relation.
If we say "no problem; just take the transitive closure of the relation"... then we get the definition we started with.
Note the key role of the zero divisors in the problem with transitivity.
