Rephrasing your question, you are asking why is the conditional d.f. of a random vector given by partial derivatives. The answer is available, for example, [**here**][1].

Just as a side note: what you have on mind is a *conditioning*. Therefore you should write
$$
\mathbb{P}\left[U_{j}\le u_{j}|U_{1}=u_{1},\ldots, U_{j-1}=u_{j-1}\right]
$$
instead of
$$
\mathbb{P}\left[U_{j}\le u_{j},U_{1}=u_{1},\ldots, U_{j-1}=u_{j-1}\right]
$$
the latter being trivially $0$, given the uniformity of all marginals.

Edit: you asked where and why did the denominator appear in (*). It is there just because the conditional distribution *is* a conditioning (by definition). You obtained the strange expression (without denominator) because you started from a wrong definition of the conditional distribution. The correct definition is the *first* expression in this answer and is computed as the limit
$$
\lim_{d_{1}\to 0,\;\ldots\;,\;d_{j-1}\to0} \mathbb{P}\left[U_{j}\le u_{j}\;|\;u_{1}\le U_{1}\le u_{1}+d_{1}\,\ldots,u_{j-1}\le U_{j-1}\le u_{j-1}+d_{j-1}\right]
$$
which boils down to
$$
\lim_{d_{1}\to 0,\;\ldots\;,\;d_{j-1}\to0} \frac{\mathbb{P}\left[U_{j}\le u_{j},u_{1}\le U_{1}\le u_{1}+d_{1}\,\ldots,u_{j-1}\le U_{j-1}\le u_{j-1}+d_{j-1}\right]}{\mathbb{P}\left[u_{1}\le U_{1}\le u_{1}+d_{1}\,\ldots,u_{j-1}\le U_{j-1}\le u_{j-1}+d_{j-1}\right]}
$$
where you can already see the denominator.

  [1]: http://en.wikipedia.org/wiki/Conditional_distribution