Yes, it does mean that. There are several ways to see this, but here is one:
If the matrix is $m\times n$, then the columns being linearly independent means the matrix has rank $n$. Thus the $m$ rows span an $n$-dimensional subspace of $\Bbb R^n$, which must be $\Bbb R^n$ itself. In particular, that means that there are linear combinations of the rows that make up each of the basis vectors.
The $k$th row of any left inverse will be the coefficients of such a linear combination for the $k$th basis vector, and any matrix consisting of such rows will be a left inverse. (In general, in a matrix product $AB=C$, the $k$th row in $C$ is a linear combination of the rows in $B$ given by the coefficients in the $k$th row of $A$. Also, more commonly, the $k$th column in $C$ will be a linear combination of the columns of $A$ given by the coefficients in the $k$th column of $B$.)