Can I get a spoiler on where this definition of the elements of a Lie group is headed?
In this lecture Alex Flournoy does a great job at introducing the elements of a Lie group. However, there is ultimately this unmotivated definition:
A general element of a Lie group can be written as
$$A = \exp \left ( g_A V^A \right)$$
where $g_A$ generates the transformation, and $V^A$ parameterizes it. So in the case of rotations in the Euclidean space, $SO(3)$ if $A$ is an $n \times n$ matrix, so is $g_A,$ while $V^A$ is an $n$ vector of parameters (angles of rotation in each plane: $xy,$ $xz$ and $yz$).
He later on explains that $g_A$ is better understood as a vector of matrices $g_A=\begin{bmatrix}R_{xy} & R_{yz} & R_{xz} \end{bmatrix},$ and the $V^A$ a vector $V^A=\begin{bmatrix}\phi&\theta&\alpha\end{bmatrix}^\top$ so that for the second entry, the Taylor series would be
$$\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\theta & - \sin \theta \\0 &\sin \theta & \cos\theta \end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1-\frac{\theta^2}2+\cdots & - \theta+\frac{\theta^3}{3!}+\cdots \\0 & \theta-\frac{\theta^3}{3!}+\cdots & 1-\frac{\theta^2}2+\cdots \end{bmatrix}$$
Is the inclusion of the exponential in the definition motivated to make the group operation (matrix multiplication) simply the addition of the exponents, as well as to show its algebraic nature by applying a Taylor expansion to the power series $\exp (x)=\displaystyle \sum_{k=0}^\infty \frac{x^k}{k!},$ with the ultimate intention of making it consistent with the definition:
A Lie group is the connected portion of a continuous group with analytic group composition function.
? Or is there a deeper reason perhaps connected to the charts of the group viewed as a manifold?
Tentative answer based on the example on this presentation by XylyXylyX on $SL(2,\mathbb R)$ parameterized as
$$\begin{bmatrix} 1+a & b \\ c & \frac{1+bc}{1+a}\end{bmatrix}$$
The generators are found through the general formula
$$X_\mu(r\times r)=\lim_{\alpha \to 0}\frac{M(0,0,\dots,\alpha^\mu,0,0)-M(0,0,\dots,0)}{\alpha^\mu}$$
which applied to the example amounts to performing the partial derivative for each parameter for the total of $3$ parameters, which is the dimension of the manifold, and evaluating the result at $(0,0,0),$ corresponding to the first term of the Taylor series (linearization):
$$\begin{align} X_a=\left. \frac{\partial}{\partial a}M(a,0,0) \right|_{(0,0,0)}= \left. \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{(1+a)^2}\end{bmatrix}\right|_{(0,0,0)} &=\begin{bmatrix} 1 & 0\\ 0 & -1\end{bmatrix}\\[2ex] X_b=\frac{\partial}{\partial b}M(0,b,0) &= \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}\\[2ex] X_c=\frac{\partial}{\partial b}M(0,0,c) &= \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix} \end{align}$$
These three generators will form a basis for a vector space, and every linear combination $\varepsilon^\alpha X_\alpha$ will be an infinitesimal generator of group elements in a given direction away from the origin in the Euclidean parametric space, made possible by the manifold structures of the group $SL(2,\mathbb R)$.
Moving away from the origin can be viewed like a process of continuous compounding these displacements in any given direction. Given that the origin in the charted Euclidean space corresponds to the identity matrix, a point in the manifold (group) along a given direction (set by the linear combination of the basis vectors derived above) will be given by
$$\begin{align} \lim_{N\to \infty} \left (\mathbf I + \frac{\varepsilon^\mu X_\mu}{N} \right )^N &= \left( \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} + \frac 1 N \left( a \begin{bmatrix} 0 & 0 \\ 0 & -1\end{bmatrix} + b \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix} \right)\right)^N\\[2ex] &= \left( \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} + \frac 1 N \begin{bmatrix} a & b \\ c & -a\end{bmatrix} \right)^N\\[2ex] &= \exp\left(\begin{bmatrix} a & b \\ c & -a\end{bmatrix}\right)\\[2ex] &=\sum_{N=0}^\infty \frac 1 {N!}\begin{bmatrix} a & b \\ c & -a\end{bmatrix}^N \end{align}$$