A wise man once told me that Lie Algebras and Lie Groups are connected in a mysterious way. I am still unsure how much of the theory I understood about it, but one particular illustrative example stuck with me:
$$\left[\begin{array}{rr} \cos(dk)&\sin(dk)\\ \sin(-dk)&\cos(dk) \end{array}\right] = \left[\begin{array}{rr} -\sin(d)&\cos(d)\\ -\cos(-d)&-\sin(d) \end{array}\right]^{k}$$
Where we recognize element-wise: $$\begin{cases}\frac{ d\cos(t)}{dt}&= -\sin(t)\\\frac{d \sin(t)}{dt} &= \cos(t)\\\frac{d\sin(-t)}{dt} &= -\cos(-t)\\\frac{d\cos(t)}{dt} &=-\sin(t)\end{cases}$$
Which I interpret as: Integrating along a curve (circle in this case) is the same as taking infinitesmal "steps" along the curves tangent. But I don't know if I may be fooling myself?
If I'm not fooling myself, how does one go from verifying this to understanding the algebra behind and move on to understand and maybe build more complicated groups / algebras for traversing routes?
All you need to understand the connection between a matrix Lie group and its Lie algebra is the matrix exponential. For any closed group $G \subset \Bbb R^{n \times n}$ (i.e. for any matrix Lie group), the corresponding Lie algebra is the set of all matrices $X$ such that for all $t \in \Bbb R$, $\exp[tX] \in G$.
Let's look at your "illustrative example" in this light: $SO(2)$ can be conveniently described as $\left\{ f(\theta): \theta \in \Bbb R \right\}$, where $$ f(\theta) = \pmatrix{\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta} $$ and the identity occurs at $\theta = 0$. The Lie algebra is the "tangent space at the identity". Note that for any $\theta(t)$ satisfying $\theta(0) = 0$, we have $$ \frac d{dt} \left. f(\theta(t))\right|_{t = 0} = \theta'(0) \pmatrix{0 & -1\\1 & 0} $$ So, the Lie Algebra is the space consisting of the span of all multiples of $\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$. You can verify that this collection satisfies our other definition too: $e^{tX}$ will only be a rotation for all $t$ if $X$ is in this "tangent space".
In particular, note that $\exp\left[t\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)\right] = \left(\begin{smallmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{smallmatrix}\right)$.
So, where do the infinitessimal steps come in? Recall that $$ \exp(X) = \lim_{n \to \infty} \left(I + \frac{X}{n}\right)^{n} $$ Intuitively: by taking an infinitessimal perturbation of the identity along the tangent space and applying it infinitely many times, we end up tracing out a path along the group. In fact, this path is a one-parameter subgroup.