A holomorphic function is conformal

The angle between two curves through a point is measured by means of the tangent vectors to the curves at the point. Then, a map preserves angles (i.e., it is conformal) if the differential of the map preserves angles. The differential of the map is the linear map defined by the Jacobian matrix. As you have said, the Jacobian corresponds to a conformal linear map, thus proving that holomorphic functions are conformal.

In the case of manifolds, a map f between two manifolds M and M’ induces a linear map between the tangent space of M at p and the tangent space of M’ at f(p), which is called the differential of f at p. The map is conformal if the differential is conformal at every point. In the present case both manifolds M and M’ are the set of complex numbers.

From $f(z)-f(z_0)=(z-z_0)f'(z_0) + o(|z-z_0|)$ we see that $f$ "hardly differs" from a multiplication with the nonzero complex number $f'(z_0)$. The little-o is really small (by definition) and lets the distinction between curves through $z_0$ of $f(z_0)$ and their tangents vanish.