Question about Do Carmo's notation Let M be a differentiable manifold. On page 26 of Riemannian Geometry by Do Carmo, he makes this statement:

Observe that if $\varphi:M\longrightarrow M$ is a diffeomorphism, $v\in T_pM$ and $f$ is a differentiable function in a neighborhood of $\varphi(p)$, we have
$$(d\varphi(v)f)\varphi(p)=v(f\circ\varphi)(p).$$
Indeed, let $\alpha:(-\varepsilon,\varepsilon)\longrightarrow M$ be a differentiable curve with $\alpha'(0)=v$, $\alpha(0)=p$. Then
$$(d\varphi(v)f)\varphi(p)=\left.\frac{d}{dt}(f\circ\varphi\circ\alpha)\right|_{t=0}=v(f\circ\varphi)(p).$$

I understand what he means by the right hand side. Namely, a vector $v$ in the tangent space at $p$ acting as a derivation on the differentiable function $f\circ\varphi$ and then evaluating the resultant function at $p$. However, I'm unsure if I'm understanding what the object on the left hand side is supposed to be. You take the differential of the map $\varphi$ (not at any point in particular?, I assume there's supposed to be a subscript $p$?) and then act this Jacobian on the vector $v$. This is fine, and it should give you a vector in the tangent space at $\varphi(p)$. You then treat this as a derivation and so it maps $f$ to some other differentiable function which you then evaluate at $\varphi(p)$. Is this correct? If so, I still can't see where the middle step comes from.
 A: Let $g = f\circ \varphi : M \to M $. Then $g$ is differentiable and the chain rules says
$$
\mathrm{d}g (p) = \mathrm{d}(f\circ \varphi)(p) = \mathrm{d}f(\varphi(p))\circ\mathrm{d}\varphi(p).
$$
If $\alpha$ is a path in $M$ with $\alpha(0)=p$ and $\alpha'(0)=v$, then by definition
$$
(g\circ\alpha)'(0) = \mathrm{d}g(\alpha(0))\alpha'(0)=\mathrm{d}(f\circ \varphi)(p)v = v\cdot \left( f\circ \varphi\right),
$$
where the last equality is just a notation for the vector $v$ acting on the function $f\circ \varphi$ by evaluating its differential.
But if you write $\beta = \varphi\circ \alpha$, which is a path with $\beta(0) = \varphi(p)$ and $\beta'(0) = \mathrm{d}\varphi(p)v$
$$
(f\circ\beta)'(0) = \mathrm{d}f(\beta(0))\beta'(0) = \mathrm{d}f\left(\varphi(p)\right) \left(\mathrm{d}\varphi(p)v\right) = \left(\mathrm{d}\varphi(p)v\right)\cdot f,
$$
where the last equality is the vector $\mathrm{d}\varphi(p)v$ acting on the function $f$.
The equality, $g \circ \alpha = f \circ \beta$ shows that
$$
v\cdot\left(f\circ \varphi \right) = \left(\mathrm{d}\varphi(p)v \right)\cdot f.
$$
A: I think do Carmo's equation does not make sense.
do Carmo introduces tangent vectors at $p \in M$ as tangent vectors at $t = 0$ of some curce $\alpha : (-\epsilon, \epsilon) \to M$ with $\alpha(0) = p$, i.e. as $\alpha'(0)$. Here $\alpha'(0)$ is defined as the function $\alpha'(0) : \mathcal D_p(M) \to \mathbb R, \alpha'(0)(f) = (f \circ \alpha)'(0)$. Note that $\mathcal D_p(M)$ is the set of functions $f : M \to \mathbb R$ which are differentiable at $p$ so that $f \circ \alpha : (-\epsilon, \epsilon) \to \mathbb R$ is a function which is differentiable at $0$ in the sense of elementary calculus and $(f \circ \alpha)'(0)$ is its usual derivative.
If $\varphi : M \to N$ is differentiable, do Carmo introduces the differential at $p$ as follows: Each $v  \in T_pM$ has the form $v = \alpha'(0)$ as above and he defines
$$d\varphi_p : T_pM \to T_{\varphi(p)}N, d\varphi_p(v) = (\varphi \circ \alpha)'(0) .$$
The claim on p.26 occurs in section 5. "Vector fields, brackets. Topology of manifolds". The equation
$$(d\varphi (v) f)\varphi(p) = v(f \circ \varphi)(p)$$
does not make sense as it stands. We have $v \in  T_pM$ and $f$ differentiable in a  neighborhood of $\varphi(p)$. Then $f \circ \varphi$ is differentiable in a  neighborhood of $p$ and $v(f \circ \varphi) \in \mathbb R$. It remains unclear what $v(f \circ \varphi)(p)$ should be. The same issue is on the LHS. However, if we omit the arguments $\varphi(p)$ and $p$, we get
$$d\varphi (v) f =  v(f \circ \varphi) .$$
It seems that the LHS stands for $d\varphi_p(v)(f)$. But then, if $v = \alpha'(0)$ for some curve $\alpha$ through $0$, we have by the above definitions $d\varphi_p(v) = (\varphi \circ \alpha)'(0)$ and
$$d\varphi_p(v)(f)  = (\varphi \circ \alpha)'(0)(f) = (f \circ \varphi \circ \alpha)'(0) = \frac{d}{dt}(f \circ \varphi \circ \alpha) \mid_{t=0}$$
and
$$(f \circ \varphi \circ \alpha)'(0) = \alpha'(0)(f \circ \varphi) = v(f \circ \varphi) .$$
Therefore
$$d\varphi_p(v)(f) = \frac{d}{dt}(f \circ \varphi \circ \alpha) \mid_{t=0} = v(f \circ \varphi)$$
which makes sense. The formula
$$d\varphi_p(v)(f) = v(f \circ \varphi)$$
is correct for any differentiable $\varphi : M \to N$; it is not limited to diffeomorphisms $\varphi : M \to M$.
Update:
As Moishe Kohan says in a comment, it is poor notation and makes sense if we give it the correct interpretation.
Indeed we have to consider a vector field $X$ on $M$ instead of a single $v \in T_pM$. Then the equation reads as
$$d\varphi(X)(f)(\varphi(p)) = X(f \circ \varphi)(p) .$$
Let $X_p = X(p) = v \in T_pM$. Then we have $X(f \circ \varphi)(p) = X_p(f \circ \varphi) = v(f \circ \varphi)$ for the RHS. On the LHS we have, as will be explained below, $d\varphi(X)(f)(\varphi(p)) = d\varphi_p(X_p)(f) = d\varphi_p(v)(f)$.
The context is this: do Carmo states on p. 25 that a vector field $X  :M \to TM$ can also be regarded as function
$$X : \mathcal D (M) \to \mathcal F (M)$$
where $\mathcal F (M)$ denotes the set of functions $M \to \mathbb R$ and $\mathcal D (M)$ is the subset of differentiable functions. For the sake of proper distinction let us temporarily  write $\bar X : \mathcal D (M) \to \mathcal F (M)$ for this alternative representation of $X$.
Then $X : M \to TM$ is differentiable iff $\bar X(\mathcal D (M)) \subset \mathcal D (M)$. Equation (5) can be written without using a coordinate parameterization as
$$\bar X(f)(p) = X(p)(f)  =X_p(f) .$$
A diffeomorphism $\varphi : M \to N$ induces isomorphisms $\varphi^* : \mathcal D (N)  \to \mathcal D (M)$ and $\varphi^* : \mathcal F (N)  \to \mathcal F (M)$ given by $\varphi^*(f) = f \circ \varphi$. Thus it assigns to each vector field $\bar X$ a new vector field $\bar X^\varphi$ on $N$ given by
$$\mathcal D (N) \stackrel{\varphi^*}{\to}  \mathcal D (M)  \stackrel{\bar X}{\to} \mathcal F (M) \stackrel{(\varphi^*)^{-1}}{\longrightarrow} \mathcal F (N) .$$
This means
$$\bar X^\varphi(f) = \bar X(f \circ \varphi) \circ \varphi^{-1} .$$
For $q = \varphi(p) \in N$ this reads as
$$\bar X^\varphi(f)(\varphi(p)) = \bar X(f \circ \varphi)(p) .$$
This is do Carmo's equation and he writes $\bar X^\varphi = d\varphi(X)$ because, as we have seen above, $\bar X^\varphi(f)(\varphi(p)) = \bar X(f \circ \varphi)(p) = \bar X_p(f \circ \varphi) = d\varphi_p(X_p)(f)$ with $d\varphi_p(X_p) \in T_{\varphi(p)}N$.
