The upshot of this answer is as follows: if a path satisfies the Euler-Lagrange equations for $L^2/2$, then it will satisfy the Euler-Lagrange equations for $L$, but the converse does not hold unless the path has affine parameterization.
Let $L = L(x, \dot x)$ be a lagrangian that is a local function of only position and velocity, then a parameterized path $x(s) = (x^i(s))$ on $M$ is said to satisfy the Euler-Lagrange equations for $L$ provided
\begin{align}
\frac{\partial L}{\partial x^i}(x(s), \dot x(s)) - \frac{d}{ds}\frac{\partial L}{\partial \dot x^i}(x(s), \dot x(s)) = 0
\end{align}
for all $i$ and for all $s$ in the domain of $x$.
Lemma 1. If $x$ satisfies the Euler-Lagrange equations for $L$, then the Beltrami Identity holds for $x$:
$$
\frac{d}{ds}L(x(s), \dot x(s)) = \frac{d}{ds}\left(\frac{\partial L}{\partial \dot x^i}\big(x(s), \dot x(s)\big)\cdot \dot x^i(s)\right)
$$
for all $s$ in the domain of $x$.
Proof. Try it yourself! The proof hinges on the fact that $L$ is a local function of only $x$ and $\dot x$.
Lemma 2. If $L(x,\dot x) = \sqrt{g_{ij}(x)\dot x^i\dot x^j}$, then $L$ satisfies the following identity:
$$
\frac{\partial (L^2/2)}{\partial \dot x^i}(x, \dot x) \dot x^i = L(x,\dot x)^2
$$
Proof. Try this yourself too! If you are stuck, you can check out a sketch of a proof below
\begin{align} \frac{\partial}{\partial \dot x^i}\left(\frac12g_{\alpha\beta}\dot x^\alpha\dot x^\beta\right) \dot x^i &= \frac12\dot{x}^i g_{\alpha\beta}\frac{\partial}{\partial\dot{x}^i}\left(\dot x^\alpha\dot x^\beta\right) \\ &= \frac12\dot{x}^i \left[g_{\alpha i}\dot{x}^\alpha+g_{i\beta}\dot{x}^\beta\right] \\ &= \frac12\dot{x}^i \left[2g_{\alpha i}\dot{x}^\alpha\right] \\ &= g_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta \end{align}
To get the third equality, use the fact that metric is symmetric.
Corollary. If $L(x,\dot x) = \sqrt{g_{ij}(x)\dot x^i\dot x^j}$, and $x$ satisfies the Euler-Lagrange equations for $L^2/2$, then $x$ satisfies the Euler-Lagrange equations for $L$.
Proof. If $x$ satisfies the Euler-Lagrange equations for $L^2$, then Lemma 1 gives the following Beltrami identity (we use notational shorthand here -- all expressions should be evaluated on $x(s)$)
$$
\frac{d(L^2/2)}{ds} = \frac{d}{ds} \frac{\partial (L^2/2)}{\partial \dot x^i}\cdot \dot x^i
$$
On the other hand, evaluating both sides of Lemma 2 on $x(s)$, and taking the derivative of both sides with respect to $s$ gives
$$
\frac{d}{ds} \frac{\partial (L^2/2)}{\partial \dot x^i}\cdot \dot x^i = \frac{d(L^2)}{ds}
$$
Combining these facts shows that $d(L^2)/ds = 0$ which implies that $L^2$ is constant along $x(s)$ and therefore that $L$ is also constant along $x(s)$:
$$
\frac{dL}{ds} = 0.
$$
Now, we separately notice that since $x$ satisfies the Euler-Lagrange equations for $L^2/2$, we have
\begin{align}
0
&= \frac{\partial(L^2/2)}{\partial x^i} - \frac{d}{ds} \frac{\partial (L^2/2)}{\partial \dot x^i} \\
&= L\left(\frac{\partial L}{\partial x^i} - \frac{d}{ds}\frac{\partial L}{\partial \dot x^i}\right) - \frac{dL}{ds}\frac{\partial L}{\partial \dot x^i} \tag{$\star$}\\
&= L\left(\frac{\partial L}{\partial x^i} - \frac{d}{ds}\frac{\partial L}{\partial \dot x^i}\right)
\end{align}
and therefore as long as $L\neq 0$, we see that $x$ satisfies the Euler-Lagrange equations for $L$ as was desired.
The crucial point here is that because of the specific form of $L$, any path satisfying the Euler-Lagrange equation for $L^2/2$ has the nice property that $dL/ds = 0$ along the path. This allows one to kill the term in $(\star)$ which is the term that is the essential difference between the Euler-Lagrange equations for $L^2/2$ and the Euler-Lagrange equations for $L$.
However, if $x$ satisfies the Euler-Lagrange equations for $L$, then it is not necessarily the case that $dL/ds = 0$ along $x$, so in this case, one can't kill that term in $(\star)$, so it need not be a solution to the Euler-Lagrange equation for $L^2/2$.
Nonetheless, if $x$ is affinely parameterized, then it will automatically have the property that $L$ is constant along it, so it will automatically satisfy both Euler-Lagrange equations.
In fact, using parts of the computations above, it is not hard to show that
Proposition. Let $L(x, \dot x) = \sqrt{g_{ij}(x)\dot x^i\dot x^j}$. A path $x$ is an affinely parameterized geodesic if and only if it solves the Euler-Lagrange equations of both $L$ and $L^2/2$.
So the Euler-Lagrange equations of $L^2/2$ yield all affiniely parameterized geodesics, while the Euler-Lagrange equations of $L$ yield all geodesics, regardless of parameterization.
