Take $x,y$ the global coordinates, $x^\prime,y^\prime$ the local coordinates, $(h,k)$ the global coordinates of the local system's origin, and $\varphi$ the anticlockwise angle from the global system's horizontal ($x$) to the local system's "horizontal" ($x^\prime$).
What is needed, then, to convert from $(x^\prime,y^\prime)$ to $(x,y)$ is to 1. shift the origin; and 2. rotate by $\varphi$ clockwise. This sequence is most easily expressed in matrix-vector notation:
$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\cos\,\varphi&\sin\,\varphi\\-\sin\,\varphi&\cos\,\varphi\end{pmatrix}\cdot\left(\begin{pmatrix}x^\prime\\y^\prime\end{pmatrix}-\begin{pmatrix}h\\k\end{pmatrix}\right)$$
or explicitly,
$$\begin{align*}x&=(x^\prime-h)\cos\,\varphi+(y^\prime-k)\sin\,\varphi\\y&=-(x^\prime-h)\sin\,\varphi+(y^\prime-k)\cos\,\varphi\end{align*}$$
Look at the following graphics
the unit vector of the spherical coordinates are the unit vector tangent to the lines where two coordinate are constants and the third change.
For example, if you only change $\theta$, with $r=2m$ and $\phi=\pi/2$ fixed, you obtain the quarter of circle in the graphics, so $e_\theta$ is tangent to this circle.
If you only change $\phi$, with $r=2m$ and $\theta=\pi/2$ fixed, you obtain the circle that appears as an ellipse in the graphics, so $e_\phi$ is tangent to this circle.
If you only change $r$, with $\phi=\pi/2$ and $\theta=\pi/2$ fixed, you obtain the positive $y$ axis, so $e_r$ is tangent to this line.
You can see that the vector $r=2me_y$ has component only along $e_r$, so the result of the problem.
In general, the unit vectors in a generic position are given by
\begin{align}
e_r &= \sin\theta(e_x\cos\phi+e_y\sin\phi)+e_z\cos\theta, \\
e_\theta &= \cos\theta(e_x\cos\phi+e_y\sin\phi)-e_z\sin\theta, \\
e_\phi &= -e_x\sin\phi+e_y\cos\phi,
\end{align}
and conversely
\begin{align}
e_x &= \cos\phi(e_r\sin\theta+e_\theta\cos\theta)-e_\phi\sin\phi, \\
e_y &= \sin\phi(e_r\sin\theta+e_\theta\cos\theta)+e_\phi\cos\phi, \\
e_z &= e_r\cos\theta-e_\theta\sin\theta.
\end{align}
If you put $\theta=\phi=\pi/2$ in these equations, you get the $e_y=e_r$ in that point.
Best Answer
When writting your new position vector $l\vec{x}$, $\vec{x}$ is unitary and dimensionless.
Let $\vec{r}$ be the position vector on your initial global coordinate system and $\vec{r}_{pos}$ be the (fixed) position of the attaching point in that same coordinate system. Then
$$\vec{r}=\vec{r}_{pos}+\vec{r}-\vec{r}_{pos}=\vec{r}_{pos}+||\vec{r}-\vec{r}_{pos}||\frac{\vec{r}-\vec{r}_{pos}}{||\vec{r}-\vec{r}_{pos}||}=\vec{r}_{pos}+l\frac{\vec{r}-\vec{r}_{pos}}{l}$$
where I used the fact that $||\vec{r}-\vec{r}_{pos}||=l$ .
You now define a new position vector $$\vec{r}^*=\vec{r}-\vec{r}_{pos}=l\frac{\vec{r}-\vec{r}_{pos}}{l}$$
$\frac{\vec{r}-\vec{r}_{pos}}{l}$ is what you called $\vec{x}$ and it's dimensionless since both $\vec{r}-\vec{r}_{pos}$ and $l$ have units of distance.