As in the other answer a vectorial boolean function with $m$ coordinates $f : \{0,1\}^n\to\{0,1\}^m$ are viewed as $m$ boolean functions $f_i : \{0,1\}^n\to\{0,1\}$.
However, it is not enough to consider $f$ to be balanced when the $f_i$ are all balanced.
For example consider $m=2,$ and $f_i(x_1,x_2,x_3)=x_1\oplus x_2,$ for $i=1,2.$
The truth table for $f_i$ is $[01100110]$ in the natural order. However, the function $f(x_1,x_2)=(f_1,f_2)$ does not have a property that we would call balanced since its truth table would be $[00,11,11,00,00,11,11,00].$
A natural requirement for a balanced vectorial boolean function with $m$ output coordinates is that all $2^m$ output $m-$tuples occur equally often in its truth table.
One way to ensure this holds is to stipulate that for all nonzero vectors $(a_1,\ldots,a_m)$ in $\{0,1\}^m,$ the boolean function $\langle a,f \rangle=a_1 f_1\oplus a_2 f_2 \oplus a_m f_m$ is itself balanced.
So if we had $f_1=x_1,f_2=x_1\oplus x_2,$ in the above example our function would be balanced with truth table $[00,11,01,10,00,11,01,10]$.
Usually we require other properties in addition to being balanced, such as high algebraic immunity, correlation immunity [=resilience if balanced], high nonlinearity, etc.
There is a good survey by Claude Carlet easily discovered online entitled "vectorial boolean functions". Early papers by Nyberg gave examples of nice balanced vectorial functions. Clearly if $m=n,$ and the function is balanced, you a nice $m\times m$ S-box, since each output pattern occurs exactly once and we have a permutation.
