Your first example may not be so illustrative. Since
$$
\begin{pmatrix} 1 & 0 & a_1 \\ 0 & 1 & a_2 \\ 0 & 0 & 1 \end{pmatrix}
\begin{pmatrix} 1 & 0 & x_1 \\ 0 & 1 & x_2 \\ 0 & 0 & 1 \end{pmatrix}=
\begin{pmatrix} 1 & 0 & a_1+x_1 \\ 0 & 1 & a_2+x_2 \\ 0 & 0 & 1 \end{pmatrix}
$$
your Lie group is isomorphic to $\mathbb{R}^2$ under addition. But if you think of those two positions as coordinates $y_1$ and $y_2$, then
\begin{align*}
\frac{\partial y_1}{x_1} &= 1 &\frac{\partial y_1}{x_2} &= 0\\
\frac{\partial y_2}{x_1} &= 0 &\frac{\partial y_1}{x_2} &= 1
\end{align*}
So for each $(a_1,a_2)$, the map $dL_{(a_1,a_2)}$ is the identity map.
A better example might be
\begin{align*}
G &= SU(2) = \left\{\begin{bmatrix} \alpha & -\bar\beta \\ \beta & \bar\alpha \end{bmatrix}
: \alpha,\beta\in \mathbb{C},\ |\alpha^2| + |\beta^2| = 1\right\} \\
\mathfrak{g} &= \mathfrak{su}(2)
= \left\{\begin{bmatrix} ix & -\bar\beta \\ \beta & -ix \end{bmatrix}
: x\in\mathbb{R},\beta\in \mathbb{C}\right\}
\end{align*}
Let $\xi = \begin{pmatrix} i & 0 \\ 0 &-i\end{pmatrix} \in \mathfrak{g}$. Then $\xi$ is the derivative at $t=0$ of the path in $SU(2)$
$$
\gamma(t) = \exp\begin{pmatrix} it & 0 \\ 0 &-it\end{pmatrix}
= \begin{pmatrix} e^{it} & 0 \\ 0 &e^{-it}\end{pmatrix}
$$
So for $g = \begin{pmatrix} \alpha & -\bar\beta \\ \beta & \bar\alpha \end{pmatrix} \in G$, we have
\begin{align*}
L_g(\xi)
&= \left.\frac{d}{dt} \begin{pmatrix} \alpha & -\bar\beta \\ \beta & \bar\alpha \end{pmatrix}
\begin{pmatrix} e^{it} & 0 \\ 0 & e^{-it} \end{pmatrix}\right|_{t=0} \\
&= \left.\frac{d}{dt} \begin{pmatrix} \alpha e^{it} & -\bar\beta e^{-it} \\ \beta e^{it} & \bar\alpha e^{-it} \end{pmatrix}\right|_{t=0} \\
&= \left.\begin{pmatrix} \alpha e^{it}(i) & -\bar\beta e^{-it}(-i) \\ \beta e^{it}(i) & \bar\alpha e^{-it}(-i) \end{pmatrix}\right|_{t=0} \\
&= \begin{pmatrix} \alpha i & \bar\beta i \\ \beta i & -\bar\alpha i\end{pmatrix}\\
&= \begin{pmatrix} \alpha & -\bar\beta \\ \beta & \bar\alpha \end{pmatrix}
\begin{pmatrix} i & 0 \\ 0 & -i\end{pmatrix}=g\xi\\
\end{align*}
Also, I'm not sure what your pictures are supposed to represent.
