Why the maximal integral curve equality $\alpha_X(s+t) = \alpha_X(s) \, \alpha_X(t)$ is true? In here and for a vector in the tangent space of a smooth manifold at the identity $X\in T_eM$ with an integral curve $\alpha_X$ along a left-invariant vector field determined by $X$, i.e. $v_X,$ with the initial point being the identity element $e$ the following equality is put forward:
$$\alpha_X(s+t) = \alpha_X(s)  \alpha_X(t)\quad, \forall x,y \in \mathbb R$$
with domain $I_{\alpha_{X}}=\mathbb R.$
The question is about the operation $\alpha_X(s)  \alpha_X(t)$ ("apply alpha $X$ of $t$, and multiply times alpha X of x"), which is later said to be "the left translation by $l_{\alpha_X(s)}$ - which is a diffeomorphism applied to alpha $X$ evaluated at $t$, i.e. $l_{\alpha_X(s)} \circ \alpha_{X}(t).$"
I see that $l_{\alpha_X(s)}$ is a diffeomorphism, but what does the circle really stand for? If it is a function composition as in $\alpha_X(\alpha_X(t))$ the input for the outer function should be time (a real number), not a point on the manifold!
What does the multiplication (element-wise? Is the multiplication the group operation?) of two curves (presumably a tuple of values for each curve in local coordinates) represent? Why doesn't it result in an entirely new, different curve rather than a different point along the curve at $s+t$? What is really going on in that multiplication of integral curves? If the explanation could be conceptual without compromising mathematical rigor, it would be greatly appreciated. I think it has to be really easy because it doesn't trigger any questions in the audience in the linked video, but I can't see what this really means. If it is embarrassingly simple, a hint is good enough, and I can delete the question.

Possible answer: the integral curve is the solution to the differential equations of a left-invariant vector field, and the different parameterization $s$ or $t$ is just a diffeomorphism on a copy of the manifold (???). The product $\alpha_X(s)  \alpha_X(t)$ is the operation on the Lie group. Much like the vector field does not change at one spot  when pushed forward by the multiplication on the left by an element of the group the corresponding point on the curve doesn't change from $\alpha_{X}(s+t)$ when the initial $\alpha_X(t)$ is left-multiplied by $\alpha_X(s)$ in the RHS of the equation. I know, I'm an idiot, and I shouldn't even be trying to have curiosity, but that's what happens in open forums - sorry for the atrocities I likely committed against these concepts.
 A: In the absence of an expert answer, I will at least summarize what I gathered so far.
The idea is that $\alpha_X$ is a homomorphism from the group of real numbers under addition $(\mathbb R, +)$ to the Lie group $G$:
$$\alpha_X(s\color{blue} + t) = \alpha_X(s) \color{red}\cdot \alpha_X(t)\quad, \forall s,t \in \mathbb R$$
where the $\color{blue}+$ is the operation in $(\mathbb R, +)$, and the $\color{red}\cdot$ the operation in the Lie group.
This is a one-parameter subgroup of $G$, which would extend to other functions, and exemplified by the exponential function of parameterized matrices with some parameter $t\in \mathbb R.$
This can be used in the proof for $\alpha_X$ as the maximum integral curve of left-invariant vector fields as follows:
$$\begin{align}
\alpha'(t) &= \frac d{dt}\alpha(t) \\[2ex]
&=\left. \frac d{ds}\right|_{s=0}\alpha(t+s) \\[2ex]
&= \left. \frac d{ds}\right|_{s=0} \alpha(t) \, \alpha(s) \\[2ex]
&\underset{*}{=} \left. \frac d{ds}\right|_{s=0} l_{\alpha(t)} \, \circ \alpha(s) \\[2ex]
&\underset{**}{=} T_e(l_{\alpha(t)})\,\alpha'(s=0)\\[2ex]
&= V_{\alpha'(0)}\,\alpha(t)
\end{align}$$
The final line being the definition of left invariant vector field associated to $\alpha'(0)$ at $\alpha(t)$. The derivations is here, and the notation, here.

$*$ Both $\alpha(t)$ and $\alpha(s)$ are points in the manifold (Lie group). Left multiplying $\alpha(s)$ by $\alpha(t)$ is a left-translation map by the point that is left-multiplying, i.e. $\alpha(t).$
$**$ Chain rule. The tangent map is at the identity element because $\alpha(s=0)$ is the identity element by the definition of integral curves.
