I feel like I didn't fully understand what an atlas of adapted charts really is. For example let's take $\mathbf{G}=SL(2,\mathbb{R})$ as a submanifold of $\mathbb{R}^4$ with the standard differential structure. Knowing this, from my understanding, $\forall p\in \mathbf{G}, \ \exists (U,\varphi) $ charts for $\mathbf{G}$ containing $p$, and $(V,\psi)$ charts for $\mathbb{R}^4$ containing $U$ such that the map $\ \psi \circ\varphi^{-1}:\varphi(U)\to\psi(V)$ has the expression $(x_1,x_2,x_3)\mapsto(x_1,x_2,x_3,0)$.
My issue with that is that I don't really grasp these charts. Concretely, $\mathbf{G}$ has dimension $3$, this means I should have charts from opens in $\mathbb{R}^3$. This makes perfectly sense as I have that elements of $\mathbf{G}$ are $4$ dimensional vectors with one of the entries linearly dependent. My problem is that how do I define those charts? My first guess is to divide $\mathbf{G}$ in to opens that cover it, let's call them $U:=\{\begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}\in\mathbf{G} : x_1 \neq 0\}, V:=\{\begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}\in\mathbf{G} : x_2 \neq 0\}$. These cover the manifold completely. If I take the inclusions in $\mathbb{R}^3$ that map $u\in U$ to $(x_1,x_2,x_3)$ and $v \in V$ to $(x_1,x_2,x_4)$ are these valid charts?
Usually, people construct such kind of charts and prove the manifold structure by the regular value theorem https://en.wikipedia.org/wiki/Preimage_theorem.
In your case, you can consider the determinant function $$ \det: M_{2\times 2}({\mathbb R}) = {\mathbb R}^4\to {\mathbb R};\ \begin{pmatrix} x_1 & x_2\\ x_3 & x_4 \end{pmatrix} \mapsto x_1x_4-x_2x_3, $$ and $SL(2, {\mathbb R})=\det^{-1}(1)$ is the inverse image of 1.
Since 1 is a regular value of $\det$ (the Jacobian matrix $(x_4, -x_3, -x_2, x_1)\neq 0$ on $\det^{-1}(1)$), so the above theorem applies and $SL_2({\mathbb R})$ is seen as a manifold.
Of course, this regular value theorem comes from the implicit function theorem, and it actually concretely gives you a way to write some charts, and even adapted charts. By the full-rank condition of regular value, you can find a nondegenerate submatrix of the Jacobian matrix, and that means that those corresponding variables can be locally solved using the other variable, by your inverse image condition. Then the other variables are your local coordinates.
In your case, you very correctly picked two open sets $U$ and $V$ on $SL(2, {\mathbb R})$.
Now let's look at $U$ where $x_1\neq 0$. This means from $$\det=x_1x_4-x_2x_3=1,$$ you can solve $x_4$ (the variable for the nondegenerate submatrix of Jacobian) as $$ x_4 = \frac{1+x_2x_3}{x_1}. $$ Then your coordinate chart is $$ \phi: U\to \phi(U)\subset {\mathbb R}^3; \begin{pmatrix} x_1 & x_2\\ x_3 & x_4 \end{pmatrix}=\begin{pmatrix} x_1 & x_2\\ x_3 & \frac{1+x_2x_3}{x_1} \end{pmatrix}\mapsto (x_1, x_2, x_3). $$
The same happens for your $V$, and you correctly spotted the map as to $(x_1, x_2, x_4)$, since $x_3$ is solved. You can even check they are $C^\infty$-compatible charts, as the transition functions are easy to see.
Now back to the adapted submanifold chart for $(U, \phi)$. There is corresponding chart $$ \tilde U=\{\vec x\in {\mathbb R}^4\,|\, x_1\neq 0\}, $$ but we don't use the usual chart $(x_1, x_2, x_3, x_4)$ (although we are in ${\mathbb R}^4$ already). Rather we have to adapt this with the implicit function theorem for your submanifold, and use the induced diffeomorphism guaranteed by the inverse function theorem: $$ \tilde\phi: \tilde U\to\tilde\phi(\tilde U); (x_1, x_2, x_3, x_4) \to (x_1, x_2, x_3, \det-1). $$ The Jabocian is full-size and invertible now. (I think it is OK in this case, but in general, you may have to shrink your $U$ and $\tilde U$, since the theorem only guarantees a local diffeomorphism.)
Then finally your inclusion $\iota: SL(2, {\mathbb R})\to {\mathbb R}^4$ restricted to these charts are in coordinates $$ \tilde\phi\circ \iota \circ\phi^{-1}: \phi(U)\to\tilde\phi(\tilde U); (x_1, x_2, x_3)\to \Big(x_1, x_2, x_3, \frac{1+x_2x_3}{x_1}\Big)\to (x_1, x_2, x_3, 0). $$