Calculating the Lie algebra of $SO(2,1)$

We identify the Lie algebra $\mathfrak{so}(2, 1)$ with the tangent space $T_1 SO(2, 1)$ to $SO(2, 1)$ at the identity $1$. As hinted in the question, for any $A \in \mathfrak{so}(2, 1)$ we can pick a curve $a: J \to SO(2, 1)$ such that $a'(0) = A$. (We'll see below that the existence of such curves is all we need, i.e., we don't need to write out curves explicitly.) Then, the characterizing equation of $A$ gives $$a(t)^T \eta a(t) = \eta,$$ and differentiating with respect to $t$ gives $$a'(t)^T \eta a(t) + a(t)^T \eta a'(t) = 0$$ (note that the r.h.s. is $0$, not $\eta$). Evaluating at $t = 0$ gives $$\phantom{(\ast)} \qquad A^T \eta + \eta A = 0. \qquad (\ast)$$ (By the way, up to this point we haven't used the form of $\eta$ yet, so this characterization holds just as well for any nondegenerate bilinear form $\eta$ in any finite dimension.)

Now we can write the Lie algebra explicitly simply working out the (linear) conditions determined by the above characterization. This is just a matter of writing out $(\ast)$ in components, but observe that the form of $\eta$ suggests a natural block decomposition of the Lie algebra. (Here this doesn't save so much effort, but this technique is quite useful for computing explicitly Lie algebras $\mathfrak{so}(\eta)$ for bilinear forms $\eta$ on higher-dimensional vector spaces.) Decompose a general element $A \in \mathfrak{so}(2, 1)$ as $$A = \begin{pmatrix} W & x \\ y^T & z \end{pmatrix},$$ where $W \in M(2, \mathbb{R})$, $x, y \in \mathbb{R}^2$, $z \in \mathbb{R}$. In this block decomposition, $\eta = \begin{pmatrix} \mathbb{I}_2 & 0 \\ 0 & -1\end{pmatrix}$ and $(\ast)$ becomes $$\begin{pmatrix} W^T & y \\ x^T & z\end{pmatrix} \begin{pmatrix} \mathbb{I}_2 & 0 \\ 0 & -1\end{pmatrix} + \begin{pmatrix} \mathbb{I}_2 & 0 \\ 0 & -1\end{pmatrix} \begin{pmatrix} W & x \\ y^T & z\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}.$$

Writing out separately the equation for each block imposes precisely the conditions $$W^T = -W, \qquad y = x, \qquad z = 0,$$ so, \begin{align} \mathfrak{so}(2, 1) &= \left\{ \begin{pmatrix} W & x \\ x^T & 0\end{pmatrix} : W^T = -W \right\} \\ &= \left\{ \begin{pmatrix} 0 & -w & x_1 \\ w & 0 & x_2 \\ x_1 & x_2 & 0 \end{pmatrix} \right\} \textrm{.} \end{align} (Of course, the condition on $W$ is just that $W \in \mathfrak{so}(2, \mathbb{R})$, where the bilinear form on $\mathbb{R}^2$ is just the standard one, i.e., the one with matrix representation $\mathbb{I}_2$.)