No, you can't do it in general (by the way, the formula displayed in the question is not well-formed: $r_*$ takes vector fields on $U$ to vector fields on $N$). At each point $x\in M$ the differential $df_x\: T_x M \to T_{f(x)}N$ is a monomorphism. However, if $X$ is a vector field on $N$ the vector $X_{f(x)}$ need not be in the image of $df_x$.
For example, of $\Bbb R \to \Bbb R^2$ is the inclusion of the $x$-axis, and $\Bbb R^2$ is given the constant unit vector field which points in the direction of the $y$-axis, how are you going to define a tangent vector at each point of the $x$-axis? The vectors of the vector field on $\Bbb R^2$ are not tangent to the $x$-axis, so what you wish to have is going to involve making choices (in differential geometry language, this choice is known as a connection).
Yes, you can choose a splitting $f^*TN \cong TM \oplus \nu$, where $\nu$ is the normal bundle (this amounts to choosing an inner product structure on $TN$, then you can project $X$ onto $TM$ via the splitting, but this is not canonical (it depends on the inner product).