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.
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.