I'm fairly sure what's intended is that ##v## is a vector on ##M## and ##f## is a function on ##N##. The action of ##v## on ##f## (if they were both on ##M##, which they are not) is defined by
$$v(f) \equiv df (v)$$
(or alternatively, that's the definition of ##df##, depending on which notions you've decided are more fundamental).
If ##f## is a function on ##N## given by ##f : N \to \mathbb{R} ; y \mapsto f(y)## for ##y \in N##, then the pullback ##\Phi^* f : M \to \mathbb{R}## is defined, for ##x \in M##, via
$$(\Phi^* f)(x) \equiv f(\Phi(x))$$
Now, as for your confusion about the equations. ##\Phi_* v## should be a vector field living on ##N##, and hence ##(\Phi_* v)(f)## should be a function living on ##N##. However, in your first equation, ##v(\Phi^* f)## is clearly a function on ##M##, not ##N##. Therefore, it is your second equation which is correct:
$$(\Phi_* v)(f) \equiv (\Phi^{-1})^* (v(\Phi^* f))$$
because now both sides of the equation live on ##N##. One must use ##(\Phi^{-1})^*## rather than ##\Phi_*##, because functions out of a space (as ##f## is) must be pulled back rather than pushed forward.
An interesting question arises when perhaps ##\Phi^{-1}## doesn't exist (for example, when ##M## has smaller dimension than ##N##, and ##\Phi## is an embedding). In this case ##\Phi_* v## is not defined on all of ##N##, but only on the portion of ##N## on which ##\Phi## is invertible. That is, ##\Phi_* v## is only defined on the image of ##\Phi##.