This sounds intuitively true. However, I have some counter claims:
Although transversal is defined on smooth manifolds, which implies the image of $df_x$ is smooth. But this does not say if the function $f$ itself is smooth, and the existence of $df_x$ only assumes $f$ is $\mathcal{C}^1$.
So can this implies $f$ is smooth, and therefore $f$ generally smooth as long as it transverses?
If $f$ is not smooth, then you cannot define $df$ on a local chart. As a result it is impossible to judge the rank of Jacobian and transversality would not make much sense. I suspect you confused whether $df_{x}$ is non-singular with whether $f$ is smooth at $x$.