According to the lecture notes I have available, finding the singular values of a $m\times n$ matrix $A$ is done by computing $A^TA$ and finding the square roots of the eigenvalues of the resulting matrix.
The problem is that my lecture notes don't specify wether $m\geq n$ or not, and since $A^TA$ has more entries than $A$ or less depending on whether, $m\geq n$ I am not sure if the dimensions of the matrix affect the computation at all.
Thank you in advance.
If $\lambda$ is a non-zero eigenvalue of $A^T A$, then $\lambda$ is an eigenvalue for $AA^T$. To see this, suppose $v$ is an eigenvector (of appropriate dimension). Then $v \neq 0$, and $$AA^T(Av) = A(A^T Av) = A(\lambda v) = \lambda(Av).$$ We just need to establish that $Av \neq 0$. If it were, then $$0 = A^T Av = \lambda v \implies v = 0$$ since $\lambda \neq 0$. This is a contradiction, hence $Av$ is an eigenvector of $AA^T$ corresponding to $\lambda \neq 0$.
This argument establishes that the non-zero eigenvalues of $A^TA$ and $AA^T$ correspond (replace $A$ with $A^T$ for the converse). Using a diagonalisation argument (both $A^T A$ and $AA^T$ are self-adjoint and hence diagonalisable), we can even match the multiplicities of these non-zero eigenvalues.
But, as you point out, $A^T A$ and $AA^T$ are of different dimensions, meaning the sums of the multiplicities must be different. The only possible conclusion is that the larger of these matrices has $0$ as an eigenvalue, with multiplicity equal to the difference plus the multiplicity of $0$ in the other.
So, in short, the singular values are identical except some extra $0$s.