@LeakyNun Okay, I will use inverse function theorem instead. Suppose $\iota : Z \to X$ is an embedding of manifolds such that $\iota(z) = x$. Then you can choose coordinate neighborhoods $(U, z_1, \cdots, z_k)$ around $z$ and $(V, x_1, \cdots, x_n)$ around $x$ such that $d\iota : T_z Z \to T_x X$ is conjugate to the inclusion $\Bbb R^k \to \Bbb R^k \times \Bbb R^{n-k} = \Bbb R^n$ as first $k$ coordinates.
Consider the map $\Phi : U \times \Bbb R^{n-k} \to \Bbb R^n$ given by $\Phi(z_1, \cdots, z_k, y_1, \cdots, y_{n-k}) = \Phi(z) + (0, \cdots, 0, y_1, \cdots, y_{n-k})$. $d\Phi_{(z, 0)}$ is the identity map now, so $\Phi$ has a local inverse by inverse function theorem.
Eh, you have to replace some thing now, and this is always hard for me to figure out. You want to replace $(V, x_1, \cdots, x_n)$ by a little neighborhood of $(z, 0)$ in $U \times \Bbb R^{n-k}$ on which $\Phi$ is invertible, and the coordinates on $U \times \Bbb R^{n-k}$ thereof.
Let's call that coordinate neighborhood $(V_1, z_1, \cdots, z_k, y_1, \cdots, y_{n-k})$, denoting the coordinates with deliberation because that's what they are on $U \times \Bbb R^{n-k}$ hence on $V_1 \subset U \times \Bbb R^{n-k}$ as well
Then $\iota : Z \to X$ is conjugate to the inclusion $(U_1, z_1, \cdots, z_k) \to (V_1, z_1, \cdots, z_k, y_1, \cdots, y_{n-k})$ where $U_1$ is small enough to fit inside $V_1$, say $U_1 = (U \times 0) \cap V$.
Aka any embedding - and we only used immersion in this proof - can be written in coordinates as the canonical embedding - immersion - of $\Bbb R^k$ in $\Bbb R^n$ in the first $k$ coordinates
Once that is done, let $Z \subset X$ be a submanifold. Choose coordinates around $z \in Z$ such that $Z$ is the subset of points of the form $(x_1, \cdots, x_k, 0, \cdots, 0)$ in the chart with coordinates $(x_1, \cdots, x_k, x_{k+1}, \cdots, x_n)$.
Then on that chart $Z$ is given by the equation $x_{k+1} = \cdots = x_n = 0$. Letting $g = (x_{k+1}, \cdots, x_n)$ with respect to those coordinates, $Z \cap U = g^{-1}(0)$, $U$ being your chart around $z$.
