Different probabilistic statement for Simple Uniform Hashing
They're not the same, but the one in Wikipedia is a consequence of the simple uniform hashing assumption.

Let $E$ be the event that $h(k_1)=h(k_2)$.  Let $F_x$ denote the event that $h(k_1)=h(k_2)=x$.  Note that

$$E = \bigcup_x F_x,$$

and the events $F_x$ are disjoint.  Therefore,

$$\Pr[E] = \sum_x \Pr[F_x].$$

Now under the simple uniform hashing assumption,

$$\begin{align*}
\Pr[F_x] &amp;= \Pr[h(k_1)=x \land h(k_2)=x]\\
&amp;= \Pr[h(k_1)=x | h(k_2)=x] \times \Pr[h(k_2)=x]\\
&amp;= \frac{1}{m} \times \frac{1}{m}\\
&amp;= \frac{1}{m^2}.
\end{align*}$$

Plugging in, we find

$$\Pr[E] = \sum_x \Pr[F_x] = \sum_x \frac{1}{m^2} = \frac{1}{m}$$

since there are $m$ possible values of $x$.

Therefore, if the simple uniform hashing assumption holds, then the equation you found in Wikipedia also holds.  The reverse isn't necessarily true.