OK, let me see first if what I understand is correct. So start with $\operatorname{SL}(2, \mathbb{R})$. This is literally the subset $$\{\begin{pmatrix}a & b \\ c & d\end{pmatrix} ``\in" \mathbb{R}^4 \mid a d - b c = 1\}.$$ It inherits a subspace topology from $\mathbb{R}^4$, and that is fine.
Now let me see if I got the coordinate system correctly. One of $a, b, c, d$ will be non-zero, say $a$. In that case, $d = \frac{1 + bc}{a}$. Let $\epsilon > 0$ be small enough so that $0 \not\in [a - \epsilon, a + \epsilon]$, and define $p : (a-\epsilon, a + \epsilon) \times (-\epsilon, \epsilon) \times (-\epsilon, \epsilon) \to \operatorname{SL}(2, \mathbb{R})$ by $(x, y, z) \mapsto \begin{pmatrix} x & y \\ z & \frac{1 + yz}{x}\end{pmatrix}$. This defines a coordinate map around any $\begin{pmatrix} a & b \\ c & d\end{pmatrix} \in \operatorname{SL}(2, \mathbb{R})$. I have not done the arithmetic, but I believe that considering the different cases will show that the transition maps are diffeomorphisms, and hence that what we are dealing with is a 3-manifold. There must be nicer coordinate systems, but I digress.
Now, we take the quotient of $\operatorname{SL}(2, \mathbb{R})$ by $\operatorname{SL}(2, \mathbb{Z})$. It gets the quotient topology, and that is also fine. But what about the parametrization? How do we give it local coordinates?
Actually, what does this space look like (geometrically speaking)?
There are two issues here. One is finding nice local coordinates for $SL(2,\mathbb R)$ and the other is passing to $SL(2,\mathbb R)/SL(2,\mathbb Z)$ and the question "what the space looks like".
Concerning coordinates on $SL(2,\mathbb R)$, the approach you suggest works in principle, but there is a lot of compatibility to check. It is much easier to use the group structure, i.e.~find coordinates around the identity matrix $\mathbb I$, say via $\phi:U\to V$, where $U\subset\mathbb R^3$ is open and $V\subset SL(2,\mathbb R)$ is an open neighborhood of $\mathbb I$. Then $u\mapsto \phi(u)\cdot A$ defines a local parametrization around a fixed matrix $X$. Then the chart changes are just given by multiplication by a fixed matrix which clearly is smooth (even polynomial). The classical choice for coordinates around the identity would by via the exponential map, so the parametrization would be $(a,b,c)\mapsto exp\begin{pmatrix} a & b\\ c & -a\end{pmatrix}$, where $exp$ is the matrix exponential. But any other choice (for example your's specialized to the case of the identity matrix) does the job as well.
Concerning the passage to $SL(2,\mathbb R)/SL(2,\mathbb Z)$, you have to observe that $SL(2,\mathbb Z)$ is a discrete subgroup of $SL(2,\mathbb R)$. Thus the natural projection $SL(2,\mathbb R)\to SL(2,\mathbb R)/SL(2,\mathbb Z)$ is a covering, so in particular, small open subsets of $SL(2,\mathbb R)$ are mapped diffeomorphically onto small open subsets of the quotient, so you can simply use the same local coordinates on the quotient.
The question "what the quotient looks like" cannot be answered locally, since it is a three-dimensional smooth manifold and thus locally looks like an open subset in $\mathbb R^3$. Globally, you can think of $SL(2,\mathbb R)$ as a three dimensional analog of a hyperboloid in $\mathbb R^4$. If you paramterize matrices as $A=\begin{pmatrix} x+y & z+w \\ z-w & x-y\end{pmatrix}$ then $\det(A)=x^2+w^2-y^2-z^2$, so $\det(A)=1$ is a hyperboloid. The subgroup $SL(2,\mathbb Z)$ can be thought of as the intersection of that hyperboliod with a lattice in $\mathbb R^4$, and then you form the quotient by this subgroup. So locally, the quotient again looks like the hyperboloid.