From Ted Shifrin's comment "We only need to restrict attention to a small neighborhood of $\phi(x)$ to apply the chain rule at a single point" in his answer, I'm trying to formalize it to make things clear. Could you have a check on my proof?
First, I recall related definitions to remove ambiguity.
A subset $X \subseteq \mathbb{R}^{N}$ is called a smooth $n$-dimensional manifold if $\forall x \in X$, $\exists$ a diffeomorphism $\varphi: U \to V$ such that $V$ is open in $X$, $U$ is open in $\mathbb{R}^{n}$, and $x \in V$. Then $\varphi$ is called a local parameterization of $V$. The inverse $\varphi^{-1}$ is called a local coordinate system, or chart, on $V$.
Let $X \subseteq \mathbb R^N$ be a smooth $n$-dimensional manifold and $x \in X$. Let $\varphi: U \to V$ be a local parameterization around $x \in V$. The continuous linear map $\mathrm d \varphi_{\varphi^{-1}(x)} : \mathbb R^n \to \mathbb R^N$ is the Fréchet derivative of $\varphi$ at $\varphi^{-1}(x) \in U$. The tangent space of $X$ at $x$, denoted by $T_xX$, is defined as the image of $\mathrm d \varphi_{\varphi^{-1}(x)}$, i.e., $$T_xX := \operatorname{im} \left (\mathrm d \varphi_{\varphi^{-1}(x)} \right ).$$
Theorem: Let $X \subseteq \mathbb R^N$ be a $k$-dimensional smooth manifold and $x \in X$. Then the tangent space $T_xX$ is a vector space of dimension $k$.
Proof: Let $\varphi:U \to V$ be a local parameterization around $x$. Here $U$ and $V$ are open in $\mathbb R^k$ and $X$ respectively. Also, $x \in V$. Let $\varphi^{-1}$ be the inverse of $\varphi$ and $y := \varphi^{-1} (x) \in U$. Let $\phi : V' \to \mathbb R^k$ be a differentiable extension of $\varphi^{-1}$ around $x$. Here $V'$ is open in $\mathbb R^N$ and $x \in V'$. Also, $\phi$ and $\varphi^{-1}$ agree on $V' \cap V$.
There exist $r_1, r_2>0$ such that $\mathbb B_{\mathbb R^N}(x, r_1) \subseteq V'$ and $\mathbb B_{\mathbb R^k}(y, r_2) \subseteq U$. Thanks to the differentiability and thus the continuity of $\varphi$, there exists $r_3 > 0$ such that $z \in \mathbb B_{\mathbb R^k}(y, r_3) \implies \varphi(z) \in \mathbb B_{\mathbb R^N}(x, r_1)$.
Let $r_4 := \min\{r_2, r_3\}$ and $U' := \mathbb B_{\mathbb R^k}(y, r_4)$. Notice that $U'$ is open in $\mathbb R^k$ and $y \in U'$. Also, $$\operatorname{im} ( \varphi_{\restriction U'} ) \subseteq \mathbb B_{\mathbb R^N}(x, r_1) \subseteq V' = \operatorname{dom} \phi.$$
Then the map $\Psi := \phi \circ \varphi_{\restriction U'}$ is well-defined and differentiable at $y$. Moreover, $\Psi$ is the identity map on $U'$, so $\mathrm d\Psi_y$ is the identity map on $\mathbb R^k$. By chain rule, $$\mathrm d\Psi_y = \mathrm d \phi_{\varphi (y)} \circ \mathrm d\varphi_{y}.$$
This means $\mathrm d \phi_{\varphi (y)}$ is surjective, while $\mathrm d \varphi_{\varphi^{-1}(x)} = \mathrm d\varphi_{y}$ is injective. Hence $\mathrm d \varphi_{\varphi^{-1}(x)}$ is indeed an isomorphism between $\mathbb R^k$ and the tangent space $T_xX$. This completes the proof.
Update:
Writing down the details reveals why we can not apply the same strategy to show $N=k$ (which is indeed not correct!). To succeed, we want to find an open neighborhood $A$ of $x$ in $V'$ such that $\Phi := \varphi \circ \phi_{\restriction A} = \operatorname{id}_{A}$. However, the only thing we know is $\varphi \circ \phi_{\restriction A \cap V} = \operatorname{id}_{A \cap V}$. This is because we have no information/restriction on the behavior of $\phi$ on $A \cap V^c$.
Let $A \subseteq \mathbb R^m$, $B \subseteq C \subseteq \mathbb R^n$, and $D \subseteq \mathbb R^p$. Let $f:A \to B$ and $g:C \to D$ be differentiable/smooth in this sense. With the same technique of restriction, we can show that $h := g \circ f:A \to D$ is also differentiable/smooth. Fix $x \in A$ and let $y:=f(x)$. Then there is a differentiable extension $\varphi:U \to \mathbb R^n$ of $f$ around $x$. Here $x\in U$, $U$ is open in $\mathbb R^m$, and $\varphi$ agrees with $f$ on $A \cap U$. Similarly, there is a differentiable extension $\phi:V \to \mathbb R^p$ of $g$ around $y$. Here $y \in V$, $V$ is open in $\mathbb R^n$, and $\phi$ agrees with $g$ on $C \cap V$. There exists $r>0$ such that $U' := \mathbb B_{\mathbb R^m}(x, r) \subseteq U$ and $x \in U'\implies \varphi(x) \in V$. Then $\phi \circ \varphi_{\restriction U'}$ is a differentiable extension of $h$ at $x$ and agrees with $h$ on $A \cap U'$.
