Let $F:M \to (N,\bar{g})$ be a smooth map between two smooth manifolds $M \subset \mathbb{R}^2$ and $N \subset \mathbb{R}^2$ of dimension 2, with $\bar{g}$ the Euclidean metric. From what I understand, the components of the pullback metric on $M$ by $F$ are given by:
$$ (F^*\bar{g})_{ij} = {\delta}_{kl} \frac{\partial x^k}{\partial u^i} \frac{\partial x^l}{\partial u^j} = \frac{\partial x^k}{\partial u^i}\frac{\partial x^k}{\partial u^j} = (D_ux)^T(D_ux), $$
where $F$ maps $(u^1,u^2)$ to $(x^1(u^1,u^2), x^2(u^1,u^2))$ and $D_ux$ is the jacobian matrix of $F$. I also understand from Lee's Introduction to Smooth Manifolds (p.332) that if $F$ is a diffeomorphism, then it is a Riemannian isometry between $(M,F^*\bar{g})$ and $(N,\bar{g})$, and I read that such an isometry preserves e.g. lengths and geodesics.
I am wondering under which conditions one can find a map $F$ such that a given Riemannian metric $g$ is the pullback of the Euclidean metric. More precisely, suppose there is a field of symmetric positive-definite matrices $m_{ij}$. Does the map whose jacobian matrix is given by $D_ux = m_{ij}^{1/2}$ (the exponent is understood to act on the eigenvalues of $m_{ij}$) and such that $(F^*\bar{g})_{ij} = (D_ux)^T(D_ux) = m_{ij}$, always define a smooth manifold $M$ through $F^{-1}(N)$? If it is the case, are the geodesics of $N$ mapped to geodesics of $M$, since $F$ is an isometry?
I have little background in differential geometry except from what I've read in Lee's book, so please feel free to point out any sloppy part in the question.
Thanks in advance!