Short answer: No, and in fact, the original map is not surjective either!
First off, I would like to emphasize that Paul Garrett's notes are super helpful as a quick cheat sheet, and I think this is a minor glitch (about which I already notified him). So please take none of what I write below as a slight on his wonderful work.
But here is the thing: The map described in section 2.5 of Paul's notes is actually not surjective (neither is the one in section 2.1 and possibly some more). Indeed, the group $SO(2,2)$ is not connected, while $SL_2(\mathbb{R})$ is connected, and the map he describes clearly is continuous, i.e., it preserves connectedness.
However, depending on what you want to achieve, one can fix this and then it extends to your case, too. Which I discuss in the ...
Long answer: Yes, after correcting the statement, and if your ground field contains no square root of $-1$
The map Paul constructs is surjective onto the connected component of the identity element. So you could restrict it to that, and then everything carries over. Alternatively, one can enlarge the domain to get a surjection this way. I'll discuss this in detail below. Which one you need of course depends on your application.
Before I do that, let me point you to the very nice arXiv preprint Shaul Zemel, An Algebraic Description of the Exceptional Isogenies to Orthogonal Groups, 2014, which deals with all this in much greater detail. I guess, you could stop reading this answer now and get everything from there. But perhaps what I write below is slightly more accessible? Besides, I'd already written it before I found Shaul's paper, so might as well leave it in now ...
Another note: while I was typing this answer, I was re-reading Paul Garrett's notes once again and realized that he mentions S. Zemel twice in foot notes! Now I wonder what the story there is... ah well.
Let's do some math now.
First we fix notation. Let $k$ be a field, let $V:=M_{2,2}(k)$ with basis
$$
e_0:=\begin{pmatrix}1 & \\ & 1\end{pmatrix},\quad
e_1:=\begin{pmatrix}1 & \\ & -1\end{pmatrix},\quad
e_2:=\begin{pmatrix} & 1 \\ -1 &\end{pmatrix},\quad
e_3:=\begin{pmatrix}& 1 \\ 1 &\end{pmatrix},
$$
and let $V_0$ be the subspace of of $V$ of matrices with trace $0$. Then $e_1,e_2,e_3$ form a basis of $V_0$.
Moreover, we set $SL_2^{\pm}(k) := \{ g\in GL_2(k) : \det(g) = \pm 1\}$. Then $SL_2^{\pm}(k) \cong SL_2(k)\times \{\pm 1\}$. Finally let
$$ G := \{ (g,h)\in SL_2^{\pm}(k) \times SL_2^{\pm}(k) \mid \det(g)=\det(h)\}. $$
As you already observed, the description in section 2.5 of Paul Garrett's notes is not specific to $\mathbb{R}$ and in fact it extends to a homomorphism $\phi:G\to SO_{2,2}(k)$ which maps $(g,h)$ to $v\mapsto gvh^{-1}$, preserves the form (verify!), and also has kernel of size 2. We will prove the following:
Theorem. The map $\phi:G\to SO_{2,2}(k)$ is a double cover (in particular surjective) if there is no $x\in k$ such that $x^2=-1$ (thus $\mathrm{char}(k)\neq 2$).
First we establish various auxiliary results.
Let $S$ be the stabilizer in $G$ of $e_0=I_2$, then (again: as you already observed) $(g,h)\in S$ if and only if $g=h$. So $S\cong SL_2^{\pm}(k)$. Furthermore $S$ necessarily also stabilizes the orthogonal complement $e_0^\perp=\langle e_1,e_2,e_3\rangle=V_0$. Note that $V=\langle e_0 \rangle \oplus V_0$ and $S$ acts trivially on the first summand. Thus $\phi(S)$ is contained in $T$, the stabilizer of $e_0$ inside $SO_{2,2}(k)$. Observe that $T \cong SO_{1,2}(k)$. Since $S\cong SL_2(k)$, the map $\phi_{|S}:S\to T$ is "essentially" the homomorphism $SL_2^{\pm}(k)\to SO_{2,1}(k)$ constructed as in section 2.2 of Paul's notes.
Lemma 1. $\phi:G\to SO_{2,2}(k)$ is surjective $\iff$ $\phi_0: SL_2^{\pm}(k)\to SO_{2,1}(k)$ is surjective.
Proof. Note that $G$ acts transitively on elements $x\in V$ for which $2\det(x)=\langle x,x\rangle = 2$. Hence if $a\in SO_{2,2}(k)$ then there is $g\in SL_2(k)\times SL_2(k)$ such that $e_0.a \phi(g)^{-1} = e_0$, i.e., $a \phi(g)^{-1}\in T$. So $\phi$ is surjective iff $\phi_{|S}:S\to T$ is surjective. $\square$
So we have reduced to studying the map $\phi_0$. To prove that it is surjective, it suffices to show that its image contains a generating set of the target group. This brings to mind the Cartan–Dieudonné theorem which for our purposes implies that $O_{2,1}(k)$ is generated by reflections on $V_0$ at anisotropic vectors $v\in V_0$ (as usual, the reflection maps $v$ to $-v$ and induces the identity on the orthogonal complement $v^\perp$).
To get the reflections, one can show that for $g\in V_0\cap SL_2(k)$ (i.e., $\det(g)=1$ and $\mathrm{tr}(g)=0$) the action $g.v \mapsto gvg^{-1}$ is actually "almost" a reflection at $g$: it fixes $g$ and acts as inverse on $g^\perp$. This follows readily from the following observation (see also the proof of Lemma 3, where we use Lemma 2):
Lemma 2. Let $x,y\in V_0$. Then $x\perp y$ (i.e. $\langle x,y\rangle = \mathrm{tr}(xy) = 0$) if and only if $xy=-yx$.
Proof. It is well-known that $\mathrm{tr}(xy)=\mathrm{tr}(yx)$. If $xy=-yx$ then also $\mathrm{tr}(xy)=-\mathrm{tr}(yx)$, hence $\mathrm{tr}(xy)=0$ (as $\mathrm{char}(k)\neq 2$).
For the converse direction, just check by direct computation (i.e. multiply two general $2\times 2$ matrices with trace $0$, then if the result also has trace 0 we get the claim). $\square$
Continuing from before the lemma, this means that $g.v \mapsto -gvg^{-1}$ is the reflection at $g$. To make this work, we temporarily switch to a larger domain and codomain:
Define a homomorphism $\rho:GL_2(k)\times\{\pm 1\}\to O_{2,1}(k)$ as follows: the $\pm1$ in the right factor is mapped to $\pm \mathrm{id}_{V_0}$ (i.e., scalar multiplication by $\pm 1$). And $g\in GL_2(k)$ is mapped to $v\mapsto gvg^{-1}$.
Lemma 3. $\rho$ is surjective.
Proof. By Cartan–Dieudonné it suffices to show that all reflections at anisotropic vectors $v\in V_0$ are in the image. But $v$ is anisotropic iff $\langle v,v\rangle \neq 0$ iff $\det v \neq 0$ iff $v\in GL_2(k)$. For $x\in V_0$ we have $\rho(v,-1).x = -vxv^{-1}$. Thus $\rho(v,-1).v = -v$; and for $x\in v^\perp$ we have using Lemma 2 that $\rho(v,-1).x = -vxv^{-1} = xvv^{-1}=x$. Thus $\rho(v,-1)$ is precisely the reflection at $v$. $\square$
The main theorem now follows from Lemma 1 together with the following:
Lemma 4. $\phi_0: SL_2^{\pm}(k)\to SO_{2,1}(k)$ is surjective if there is no $x\in k$ with $x^2=-1$.
Proof. The restriction of $\rho$ to $SL_2(k)\times\{1\}$ is essentially the map $\phi_0$. Since $\rho$ is surjective, for every $a\in SO_{2,1}(k)$ there is $(g,\epsilon)\in GL_2(k)\times\{\pm 1\}$ such that $\rho(g,\epsilon)=a$. Then $g = h\cdot u$, for some $h\in SL_2^\pm(k)$ and $u=\begin{pmatrix} x & 0 \\ 0 & 1 \end{pmatrix}$. We already know that $\rho(h,1)\in SO_{2,1}(k)$. Hence $\rho(g,\epsilon)=\rho((h,1)\cdot(u,\epsilon))=\rho(h,1)\cdot\rho(u,\epsilon)\in SO_{2,1}(k)$ iff $\rho(u,\epsilon)\in SO_{2,1}(k)$. We compute
$$
(u,\epsilon).e_1 = \epsilon e_1, \quad
(u,\epsilon).e_2 = \epsilon x e_2, \quad
(u,\epsilon).e_3 = \epsilon x e_3,
$$
hence $\det(\rho(u,\epsilon))=\epsilon^3 x^2$. Since $x^2\neq -1$, this determinant is equal to $\pm 1$ if and only if $\epsilon=1$ and $x=\pm 1$. The latter implies $u\in SL_2^\pm(k)$ thus $g\in SL_2^\pm(k)$. Hence $a = \rho(g,1) = \phi_0(g)$.
$\square$
Actually it is not hard to extend the argument in the proof to see that the converse also holds, i.e., if there is $x\in k$, with $x^2=-1\neq 1$, then $\rho_0$ (and hence $\rho$) is not surjective, not even if we enlarge its domain to $GL_2(k)$.
