Defining tabular rows with macros using `\ifnum` leads to `Incomplete \ifnum`

You can skip & inside the conditional, you just have to hide it a bit.

\documentclass{article}

\newcommand{\test}[3]{%
    \ifnum\numexpr#1\relax=0
        AA#2 \uppercase{&} BB#3
    \else
        CC#2 \uppercase{&} DD#3
    \fi
}

\begin{document}

    \begin{tabular}{l|l}
        \test{0}{col1-1}{col1-2} \\
        \test{1}{col2-1}{col2-2} \\
    \end{tabular}

\end{document}

You can't start a conditional in a table cell and end it in another, because TeX inserts an inaccessible token that signals the end of a cell and which is not allowed in text skipped in a conditional.

The usual trick is to do all the conditional and then execute one of the two codes:

\documentclass{article}

\makeatletter
\newcommand{\test}[3]{%
  \ifnum\numexpr#1\relax=0
    \expandafter\@firstoftwo
  \else
    \expandafter\@secondoftwo
  \fi
  {#2 & }
  {#2 & #3}
}
\makeatother

\begin{document}

\begin{tabular}{ll}
  \test{0}{col1-1}{col1-2} \\
  \test{1}{col2-1}{col2-2} \\
\end{tabular}

\end{document}

See What do \@firstoftwo and \@secondoftwo do? for more information.

In tabular environment & cannot be skipped, so when your condition is true LaTeX will do the following: it scans #2 & #3, then when it encounter \else the expansion stops, however, when the second & is met the tabular environment starts expansion again, so the output becomes #2 & #3 & #3. It can be fix by building new control sequence, for example:

\documentclass{article}
\newcommand{\testa}[2]{#1 & #2}
\newcommand{\testb}[2]{#1 & #2}
\newcommand{\test}[3]{
    \ifnum\numexpr#1\relax=0
        \testa{#2}{#3}
    \else
        \testb{#2}{#3}
    \fi
}

\begin{document}

    \begin{tabular}{ll}
        \test{0}{A}{B} \\
        \test{1}{B}{A} \\
    \end{tabular}

\end{document}

gives you: