As @InTransit suggested in a comment, O'Neill's phrase "and defined on an open subset of $\mathbb R^m$" seems to take care of the problem. Note that the subset of $\mathbb R^m$ on which $\psi\circ f \circ \varphi^{-1}$ is defined is
$(f\circ\varphi^{-1})^{-1}(V) = \varphi(f^{-1}(V)\cap U)$. O'Neill's definition stipulates that this set is open in $\mathbb R^m$. Because $\varphi$ is, in particular, a homeomorphism from an open subset of $M$ to an open subset of $\mathbb R^m$, this implies that $f^{-1}(V)\cap U$ is open in $M$. Knowing this, we can replace $U$ by $U_0 = f^{-1}(V)\cap U$ and $\varphi$ by $\varphi|_{U_0}$, and then smoothness by O'Neill's definition implies smoothness by mine.
If you omit the requirement that the composite map be defined on an open set, but only require, as the OP suggested, that $\psi\circ f \circ \varphi^{-1}$ have a smooth extension to an open neighborhood of each point, then you won't necessarily get continuity. A counterexample (taken from Problem 2-1 in my smooth manifolds book, 2nd ed.) is the function $f\colon\mathbb R\to \mathbb R$ defined by
$$
f(x) = \begin{cases}
1, &x\ge 0,\\
0, &x< 0.
\end{cases}
$$
Just to elaborate on Xiao's answer; considering the fact that we refer to tangent vectors derivations then the differential or push forward map;
$$[f_{*,p}(X_p)] g = X_p( g \circ f); \ X_p \in T_pM, g \in C_{f(p)}^{\infty}(N), f: M \to N$$
here $X_p( g \circ f) \in T_{f(p)}N$ (why?). Well if you take $g,h \in C_{f(p)}^{\infty}(N)$ then;
$$[f_{*,p}(X_p)] (gh) = X_p(gh \circ f) = X_p(g \circ f \cdot h \circ f)$$
and now since $X_p$ is a derivation;
$$X_p(g \circ f \cdot h \circ f) = X_p(g \circ f) \cdot h(f(p)) + g(f(p)) \cdot X_p(h \circ f) $$
$$ \hspace{1.2in}= [f_{*,p}(X_p)]g \cdot h(f(p)) + g(f(p)) [f_{*,p}(X_p)] h$$
The linearity piece is also clear since $X_p$ is linear. Therefore; if you take $f: M \to \mathbb{R}$ and $(U,x^1,...,x^d)$ to be a chart about $p$ then;
$$\left\{\frac{\partial}{\partial x^1}\Bigr|_p,...,\frac{\partial}{\partial x^d}\Bigr|_p\right\}$$
is a basis for $T_pM$. Similarly, we can use the coordinate $t$ to parametrize a neighborhood of $f(p) \in \mathbb{R}$ and so $T_{f(p)}\mathbb{R}$ has basis vector;
$$\frac{\partial}{\partial t}\Bigr|_{f(p)} := \frac{d}{dt}\Bigr|_{f(p)}$$
Since $f_{*,p}$ is linear, it maps tangent vectors to tangent vectors i.e;
$$f_{*,p}\left(\frac{\partial}{\partial x^i}\right) = \alpha \frac{d}{dt}\Bigr|_{f(p)}$$
If we evaluate both sides at $t$ and use the definition of the differential we have;
$$f_{*,p}\left(\frac{\partial}{\partial x^i}\right) t = \alpha \frac{d}{dt}\Bigr|_{f(p)} t \Rightarrow \frac{\partial}{\partial x^i}\Bigr|_p (t \circ f) = \frac{\partial}{\partial x^i}\Bigr|_p f = \alpha$$
The above follows from the fact that the coordinate function $t$ picks out the first coordinate of the map $f$, which is real-valued, so that if just $f$. It now follows that;
$$f_{*,p}\left(\frac{\partial}{\partial x^i}\right) =\frac{\partial}{\partial x^i}\Bigr|_p f \frac{d}{dt}\Bigr|_{f(p)}$$
Best Answer
Recall that if $\alpha$ is a curve in $M$ and $h\colon M \to \mathbb{R}$ is smooth, then the derivation associated to $\alpha'(0)$ is given by $$\alpha'(0)h = (h\circ \alpha)'(0).$$ We will use this fact twice.
As in your notation, we let $f\colon M \to N$ be a smooth map between manifolds, $p \in M$, $v \in T_pM$. Let $\gamma$ be a smooth curve on $M$ that represents $v$, and let $\mathcal{D}$ be the derivation at $p \in M$ that represents $v$. Let $g\colon N \to \mathbb{R}$ be a smooth function.
In the first definition, we have $df_p(v) = (f\circ \gamma)'(0)$, which is a tangent vector on $N$, and therefore may be regarded as a derivation at $f(p) \in N$. So, we can evaluate $df_p(v)$ at the function $g$:
$$\begin{align*} [df_p(v)](g) & = [(f\circ \gamma)'(0)](g) \\ & = (g \circ (f\circ\gamma))'(0) \\ & = ((g\circ f)\circ \gamma)'(0) \\ & = \gamma'(0)(g\circ f) \\ & = \mathcal{D}(g\circ f) \\ & = df_p(\mathcal{D})(g) \end{align*} $$
This shows that your two definitions coincide.
Edit: I realize that I have not addressed your actual question, which is whether or not your reasoning is correct. However, I think it is instructive nevertheless for you to see how to approach this problem without reference to coordinates and matrices.