Which vectors are stretched the most in a transform?

Let $U\Sigma V^*$ be the singular value decomposition of matrix $M$. Without loss of generality, assume the largest singular value is given by $\Sigma_{11}$. Let us analyze the action of $M$ on a vector $\vec v$: there is a rotation of $\vec v$ by $V^*$, followed by stretching by $\Sigma$ and then rotation by $U$. Since $\Sigma_{11}$ gives the largest stretch, the vector that gets mapped to $(1,0, 0,\dots)^T$ under $V^*$ will end up getting stretched the most and gets mapped to the major axis of the ellipse in the 2D case (after rotation by $U$). $$ V^*\vec v= (1,0,\dots)^T\implies \vec v = V (1,0,\dots)^T $$ The major axis will be $M\vec v$ where $\vec v$ is found from the above equation.

Update: OP is asking for a solution without SVU decomposition. To solve the same problem using calculus, you are looking for a vector that maximizes $$f(\vec v)=\frac{|M\vec v|^2}{|\vec v|^2}$$ We can achieve that by setting the gradient of $f$ to zero: $$ \nabla f = \nabla \left(\frac{(M\vec v)\cdot(M\vec v)}{\vec v\cdot \vec v}\right) = \frac{2M^*M\vec v}{|\vec v|^2}-\frac{2\vec v|M\vec v|^2}{|\vec v|^4}=\vec 0. $$ Since the length of $\vec v$ does not matter, I'm just going to set it to $1$. Now, we have $$ M^*M\vec v = |M\vec v|^2 \vec v. $$ This tells you that the vector maximizing $f$ is an eigenvector of $M^*M$ with eigenvalue $|M\vec v|^2$. Note that all of the eigenvectors of $M^*M$ are critical points of $f$, but only the one with the largest eigenvalue is the one that maximizes $f$.

To see how this is consistent with the solution from SVU decomposition, note that

1. If $M=U\Sigma V^*$, then $M^*M = V\Sigma U^*U\Sigma V^* = V\Sigma^2V^*$ with eigenvectors that are the right singular vectors of $M$ and eigenvalues being the square of singular values of $M$.
2. If $\vec v$ is an eigenvector of $M^*M$ with length $1$ and eigenvalue $\lambda$, $|M\vec v|^2 =\vec v.(M^*M\vec v)= \lambda \vec v.\vec v= \lambda $.