$(1)$ and $(2)$ are equivalent and are adequate definitions, but $(3)$ is too weak to characterize a submanifold.
Let us first introduce some notation.
For a smooth map $\psi : U \to V$ between open subsets $U \subset \mathbb R^m$ and $V \subset \mathbb R^p$ let us write $\psi'(x)$ for the Jacobian matrix of $\psi$ at $x$. This is what you denote as $\dfrac{d\psi}{dx} \mid_x$.
Let $\mathbb R^{k,n} = \{(x_1,\ldots,x_n) \in \mathbb R^n \mid x_{k+1} = \ldots = x_n = 0 \}$. Then $\mathbb R^{k,n} \subset \mathbb R^n$ whereas $\mathbb R^k \subset \mathbb R^n$ does not make sense (or is a serious abuse of notation).
Let us call a subset $M \subset \mathbb R^n$ a $k$-dimensional smooth submanifold of $\mathbb R^n$ of type $(1) - (5)$ if each $x \in M$ satisfies
(1) There exists a smooth $F : O_x \to \mathbb{R}^{n-k}$ defined on an open neighborhood $O_x$ of $x$ in $\mathbb R^n$ such that $\operatorname{rank} F'(x) = n-k$ and $O_x \cap M = F^{-1}(0)$.
(2) There exists a smooth $f : V_0 \to \mathbb R^n$ defined on an open neighborhood of $0$ in $\mathbb{R}^k$ such that $f(0)= x$ and $\operatorname{rank} f'(0) = k$, $f(V_0)$ is an open subset of $M$ and $f : V_0 \to f(V_0)$ is a homeomorphism.
(3) There exists a smooth $\phi : O_x \to O_0$ from an open neighborhood $O_x$ of $x$ in $\mathbb{R}^{n}$ onto an open subset $O_0 \subset \mathbb{R}^n$ such $\phi(x) = 0$, $\operatorname{rank} \phi'(x) = n$ and $\phi(O_x \cap M)= \mathbb{R}^{k,n} \cap O_0$.
(4) There exists a smooth $\phi : O_x \to \mathbb R^n$ defined on an open neigborhood $O_x$ of $x$ such that $\operatorname{rank} \phi'(x) = n$ and $\phi^{-1}(\mathbb R^{k,n}) = O_x \cap M$.
(5) There exists a diffeomorphism $h : O_x \to V$ between an open neigborhood $U_x$ of $x$ in $\mathbb R^n$ and an open subset $V \subset \mathbb R^n$ such that $h(O_x \cap M) = \mathbb R^{k,n} \cap V$.
In my opinion the most natural definition is (5). If we want, in (5) we can moreover require that $h(x) = 0$. To see this, note that $h(x) \in \mathbb R^{k,n}$. The translation $t_{-h(x)} : \mathbb R^n \to \mathbb R^n, t_{-h(x)}(y) = y - h(x)$, is a diffeomorphism such that $t_{-h(x)}(\mathbb R^{k,n}) = \mathbb R^{k,n}$, thus $h^* = t_{-h(x)} \circ h : U_x \to V^* = t_{-h(x)}(V)$ will do what we desire.
The same is true for (4): We can require that $\phi(x) = 0$.
Now let us observe that type $(3)$ submanifolds are not what we expect. As an example take $M = M_1 \times \{0\} \subset \mathbb R^2$ with $M_1 = (-\infty,1] \cup (2,\infty) \subset \mathbb R$. This satisfies $(3)$: For all $x \in M' = (-\infty,1) \times \{0\} \cup (1,\infty) \times \{0\}$ we may take $O_x = (-\infty,1) \times \mathbb R \cup (1,\infty) \times \mathbb R$, $O_0 = O_x$ and $\phi = id$. For $x = (1,0)$ we take $O_x = O_0 = \mathbb R^2$ and $\phi(x_1,x_2) = (x_1^3 - 9x_1,x_2)$. Then $\phi'(x)$ is a diagonal matrix with entries $3x_1^2-9$ and $1$ which has rank $2$ execpt for $x_1 = \pm \sqrt 3$. In particular $\operatorname{rank} \phi'(1,0) = 2$. The function $\phi_1 : \mathbb R \to \mathbb R,\phi_1(x_1) = x_1^3-9x_1$, has the property $\phi_1(x_1) \to \pm \infty$ as $x_1 \to \pm \infty$. Hence by the IVT $(-\infty,-8] = (-\infty,\phi_1(1)] \subset \phi_1((-\infty,1])$ and $(-10,\infty) = (\phi_1(2),\infty) \subset \phi_1(2,\infty)$. Hence $\phi_1(M_1) = \mathbb R$, thus $\phi$ maps $\mathbb R^2$ onto $\mathbb R^2$ and $\phi(M \cap \mathbb R^2) = \phi(M) = \phi_1(M_1) \times \mathbb \{0\} = \mathbb R \times \{0\} = \mathbb R^{1,2} = \mathbb R^{1,2} \cap \mathbb R^2$.
The problematic point is the boundary point $(1,0)$ of $M$: It prevents $M$ from being a manifold, but satisfies $(3)$.
The correct substitute for $(3)$ is $(4)$. Note, by the way, that $(5)$ trivially implies $(3)$.
Let us show the equivalence of $(1), (2), (4), (5)$.
$(4) \Rightarrow (5)$: It is well-known from multivariable calculus that $\operatorname{rank} \psi'(x) = n$ implies that that there exist open neigborhoods $U_x \subset O_x$ of $x$ and $V \subset \mathbb R^n$ of $\psi(x)$ such that $\psi(U_x) = V$ and $h = \psi \mid_{U_x}: U_x \to V$ is a diffeomorphism. Let us prove $h(U_x \cap M) = \mathbb R^k \cap V$. We have $h(U_x \cap M) = \psi(U_x \cap M) \subset \psi(O_x \cap M) = \psi(\psi^{-1}(\mathbb R^k)) \subset \mathbb R^k$, thus $h(U_x \cap M) \subset \mathbb R^k \cap V$ since $h(U_x \cap M) \subset h(U_x) = V$. Moreover $h^{-1}(\mathbb R^k \cap V) \subset \psi^{-1}(\mathbb R^k \cap V) \subset \psi^{-1}(\mathbb R^k) = O_x \cap M$, thus $h^{-1}(\mathbb R^k \cap V) \subset U_x \cap M = U_x \cap (O_x \cap M)$ since $h^{-1}(\mathbb R^k \cap V) \subset h^{-1}(V) = U_x$. Therefore $\mathbb R^k \cap V = h(h^{-1}(\mathbb R^k \cap V)) \subset h(U_x \cap M)$.
$(5) \Rightarrow (4)$: Take $\phi = i \circ h$ with inclusion map $i : V \to \mathbb R^n$. Clearly $\operatorname{rank} \phi'(x) = n$. Moreover $\phi^{-1}(\mathbb R^{k,n}) = h^{-1}(i^{-1}(\mathbb R^{k,n})) = h^{-1}(\mathbb R^{k,n} \cap V)$.
$(5) \Rightarrow (1)$: Let $p : \mathbb R^n \to \mathbb R^{n-k}$ denote the projection onto the first $n-k$ coordinates. We have $\operatorname{rank} p'(y) = n-k$ for all $y$. But then $F = p \mid_V \circ h$ has the property $\operatorname{rank} F'(q) = n-k$ for all $q \in U$. Clearly $p^{-1}(0) = \mathbb R^k$, hence $F^{-1}(0) = h^{-1}(p_V^{-1}(0)) = h^{-1}(\mathbb R^k \cap V) = U \cap M$.
$(1) \Rightarrow (4)$: The $(n-k) \times n$ matrix $F'(x)$ has rank $n-k$, thus among its $n$ column vectors $a_j$ we can pick $n-k$ linearly independent vectors $a_{j_1},\ldots,a_{j_{n-k}}$ with $1 \le j_1 \le j_2 \le \ldots \le j_{n-k} \le n$. Write $C = \{j_1,\ldots,j_{n-k}\}$ and $D = \{ 1, \ldots, n\} \setminus C = \{r_1, \ldots,r_k\}$ with $1 \le r_1 \le r_2 \le \ldots \le r_k \le n$. Define
$$\psi : O_x \to \mathbb R^n, \psi_r(x_1,\ldots,x_n) = \begin{cases} F_i(x_1,\ldots,x_n) & r = j_i \in C \\ x_r & r\in D \end{cases}$$
For $r \in D$ the $r$-th row of $\mathcal A = \psi'(x)$ has the form $(0 \ldots 0 \phantom{.} 1 \phantom{.} 0 \ldots 0)$ where the entry $1$ is at position $r$. We can compute $\det \mathcal A$ by Laplace expansion along rows. Starting with $r = r_k$ we get $\det \mathcal A = (-1)^{r_k + r_k} \det \mathcal A_{r_k r_k} = \det \mathcal A_{r_k r_k}$, where $\mathcal A_{r_k r_k}$ is obtained from $\mathcal A$ by by removing the $r_k$-th row and the $r_k$-column. Continuing this process for rows $r_{k-1},\ldots, r_1$ with end with $\det \mathcal A = \det \mathcal C$ where $ \mathcal C$ has columns $c_i = a_{j_i}$. Thus $\det \mathcal A = \det \mathcal C \ne 0$, i.e. $\operatorname{rank} \psi'(x) = n$. Now let $\sigma$ be the permutation of $\{1,\ldots,n\}$ such that $\sigma(j_i) = k + i$ for $i = 1,\ldots, n-k$ and $\sigma(r_i) = i$ for $i =1,\ldots,k$. Define $P : \mathbb R^n \to \mathbb R^n, P(x_1,\ldots,x_n) = (x_{\sigma(1)},\ldots, x_{\sigma(n)})$. This is a linear isomorphism, thus a diffeomorphism. The map $\phi = P \circ \psi$ has $\operatorname{rank} \phi'(x) = n$ and is given by
$$\phi_r(x_1,\ldots,x_n) = \begin{cases} x_r & r \in \{1,\ldots,k\} \\ F_i(x_1,\ldots,x_n) & r = k + i \end{cases}$$
Therefore $\phi^{-1}(\mathbb R^{k,n}) = \{ \xi \mid \phi(\xi) \in \mathbb R^{k,n} \} = \{\xi \mid \phi_r(\xi) = 0 \text{ for } r > k \} = \{\xi \mid F_i(\xi) = 0 \text{ for } i \le n-k \} = F^{-1}(0) = O_x \cap M$.
$(5) \Rightarrow (2)$: Let $h : O_x \to V$ as in $(5)$, where $h(x) = 0$. Let $i^{k,n} : \mathbb R^k \to \mathbb R^n, i^{k,n}(x_1\ldots,x_k) = (x_1\ldots,x_k,0 \dots, 0)$ and $V_0 = (i^{k,n})^{-1}(V)$. This is an open neigborhood of $0$ in $\mathbb R^k$ and $f : V_0 \stackrel{i^{k,n}}{\to} V \stackrel{h^{-1}}{\to} O_x$ is smooth with $f(0) = x$. We have $\operatorname{rank} (i^{k,n})'(y) = k$ for all $y$, thus $\operatorname{rank} f'(0) = k$. Clearly $f$ maps $V_0$ homeomorphically onto $O_x \cap M = f(V_0)$ which is open in $M$.
$(2) \Rightarrow (5)$: The $n \times k$ matrix $f'(0)$ has rank $k$, thus among its $n$ row vectors $b_j$ we can pick $k$ linearly independent vectors $b_{j_1},\ldots,b_{j_k}$ with $1 \le j_1 \le j_2 \le \ldots \le j_k \le n$. Write $C = \{j_1,\ldots,j_k\}$ and $D = \{ 1, \ldots, n\} \setminus C = \{r_1, \ldots,r_{n-k}\}$ with $1 \le r_1 \le r_2 \le \ldots \le r_{n-k} \le n$. Define
$$F: V_0 \times \mathbb R^{n-k} \to \mathbb R^n, F_r(x_1,\ldots,x_n) = \begin{cases} f_r(x_1,\ldots,x_k) & r \in C \\ f_r(x_1,\ldots,x_k) + x_{k+i} & r = r_i \in D \end{cases}$$
For $i =1, \ldots,n-k$ the $(k+i)$-th column of $\mathcal B = F'(x)$ has the form $(0 \ldots 0 \phantom{.} 1 \phantom{.} 0 \ldots 0)^T$ where the entry $1$ is at position $r_i$. We can compute $\det \mathcal B$ by Laplace expansion along columns. As above we get $\det \mathcal B = \det \mathcal C$ where $ \mathcal C$ has rows $c_i = b_{j_i}$. Thus $\operatorname{rank} F'(x) = n$. Therefore there exist open neigborhoods $U_x$ of $x$ in $ \mathbb R^n$ and $W$ of $0$ in $V_0 \times \mathbb R^{n-k}$ such that $F(W) = U_x$ and $g = F \mid_{U_x}: W \to U_x$ is a diffeomorphism. We can write $W \cap \mathbb R^{k,n} = W_0 \times \{0\}$ with $0 \in \mathbb R^{n-k}$ and open $W_0 \subset V_0$. We get $g(W \cap \mathbb R^{k,n}) = g(W_0 \times \{0\}) = F(W_0 \times \{0\}) = f(W_0)$. Thus $g(W \cap \mathbb R^{k,n})$ is an open subset of $M$ which is clearly contained in $U_x \cap M$, but we cannot be sure that $g(W \cap \mathbb R^{k,n}) = U_x \cap M$ since points of $W \setminus \mathbb R^{k,n}$ could be mapped to $M$. Let us resolve this problem. Since $f(W_0)$ is open in $M$, each $y \in f(W_0)$ admits an open Euclidean ball $B(y)$ with center $y$ such that $B(y) \cap f(W_0) \subset f(W_0)$ and $B(y) \subset U_x$. Then $O_x = \bigcup_{y \in f(W_0)}B(y) \subset U_x$ is an open neigborhood of $x$ in $\mathbb R^n$ such that $O_x \cap M = f(W_0)$ and $V = g^{-1}(O_x)$ is an open neigborhood of $0$ in $\mathbb R^n$ such that $g(V \cap \mathbb R^{k,n}) = f(W_0) = O_x \cap M$ and $h = g^{-1} : O_x \to V$ is the desired diffeomorphism.
