Wrong pushforward of vector field definition on wikipedia Wikipedia claims that for $$\mathrm d \varphi_x:T_xM\to T_{\varphi(x)}N\,$$
we have for $X \in T_pM$ and $f \in C^{\infty}(N,\mathbb{R})$
$$\mathrm d\varphi_x(X)(f) = X(f \circ \varphi)$$ whereas I would have assumed that
$$\mathrm d\varphi_x(X)(f) = X(f \circ \varphi) \circ \varphi^{-1}.$$
Presumably, wikipedia is correct, but I currently don't see where I could be wrong here?
EDIT (See this one for explanation)
So yes, probably I would require that $\phi$ is a diffeomorphism, but in another question someone suggested the alternative equation to me and so I got confused which one is correct.
Cause I feel that $\mathrm d\varphi_x(X)(f)$ is a map $C^{\infty}(N,\mathbb{R})$ whereas $X(f \circ \varphi)$ is a map $C^{\infty}(M, \mathbb{R})$, so something does not add up there.
 A: What does your use of $\varphi^{-1}$ mean, when the original smooth map $\varphi:M\to N$ need not have been a bijection (much less a diffeomorphism)?
As to what I suspect was your thought process, I suppose you wanted to use $\varphi^{-1}$ (which, as I've just said, may not mean anything anyway) to "bring things back" to $M$ for $X$ to act on them. However, you should note that $\mathrm{d}\varphi_x(X)(\,\cdot\,)$ as defined on Wikipedia already eats smooth functions on $N$ and outputs real numbers, there's no need to use a $\varphi^{-1}$.
A: I think everything would be clear if you think carefully what is $T_xM$: an element $X \in T_xM$ is an derivation of $\mathscr D_x$, the germs of smooth functions at $x$. It's a linear map 
$$X : \mathscr D _x \to \mathbb R. $$
So an element in $[f] \in \mathscr D_x$ is represented by a smooth function $f$ defined locally at $x$. 
Now it is claimed that $d\varphi_x$ is given by
$$d\varphi_x(X)(f) = X(f \circ \varphi).$$
This is not so precise: it should be 
$$(*)\ \ \ d\varphi_x(X)[f]= X[f \circ \varphi],$$
where $[f]\in \mathscr D_{\varphi(x)}$. Note that $f\circ \varphi$ is now a function define locally at $x$. Thus $X[f\circ \varphi]$ make sense. 
Now we try to make sense of your statement (where $f$ is defined on $N$)
$$(**)\ \ \ \ \mathrm d\varphi_x(X)(f) = X(f \circ \varphi) \circ \varphi^{-1}.$$
Fix each $y\in N$. Define the linear map 
$$[\cdot]_y : C^\infty(N) \to \mathscr D _y$$
by sending $f$ to it's equivalence class $[f]_y \in \mathscr D_y$. Then $(**)$ is really $(*)$, as $(**)$ should be written as 
$$\mathrm d\varphi_x(X)[f]_{\varphi(x)} = X[f \circ \varphi]_x.$$
If you write $m = \varphi(x)$ and $\varphi$ is a diffeomorphism, then it's the same as 
$$\mathrm d\varphi_{\varphi^{-1}(m)} (X)[f]_{m} = X[f \circ \varphi]_{\varphi^{-1}(m)}.$$
It seems that in the other answer, this is exactly want the answer-er refers to. After all it is just a variant of notations. They corresponds to the same thing. 
