The Euler-Lagrange equations are obtained by employing the principle of least action. This means that if we make a slight change to the path of the particle (call it $\delta x$(t)), the action should not change (to first order in $\delta x$). We write $\delta S$ for the change of the action, so the Euler-Lagrange equations are obtained by requiring that $\delta S = 0$ for any slight change of the path $x(t)$ with the ends held fixed.
Let us calculate the change of the action $\delta S$ (we ommit terms of higher order in $\delta x$):
$\delta S = \int_{t_0}^{t_1}L(x + \delta x,\dot x + \delta\dot x,t)\mathrm{d}t - \int_{t_0}^{t_1}L(x,\dot x,t)\mathrm{d}t = \int_{t_0}^{t_1}(L_{\,x}\delta x+L_{\,\dot x}\delta \dot x)\mathrm{d}t = \int_{t_0}^{t_1}(L_{\,x} - \frac{\mathrm{d}}{\mathrm{d}t} L_{\,\dot x})\delta x \; \mathrm{d}t + \left.L_{\,\dot x}\delta x\right|_{t_0}^{t_1}$
In the last step we integrated by parts. Now the last term can be omitted because the ends of the path are held fixed, so $\delta x(t_0) = \delta x(t_1) = 0$. Since $\delta S = 0$ should hold for otherwise arbitrary $\delta x$ we conclude that
$L_{\,x} - \frac{\mathrm{d}}{\mathrm{d}t} L_{\,\dot x} = 0$
which is precisely the Euler-Lagrange equation.
Now the problem with your Lagrangian is that it is essentially zero. This can be seen by noting that I can add a total time derivative of an arbitrary function $\frac{\mathrm{d}}{\mathrm{d}t}g(x,t)$ to a Lagrangian without changing its physical meaning. This added term would amount to the expression $g_x\delta x|_{t_0}^{t_1}$ in the change of the action and therefore give zero, since again the ends of the path are held fixed.
Since any two Lagrangians are equivalent if they differ by a total time derivative $\frac{\mathrm{d}}{\mathrm{d}t}g(x,t)$ and yours is obtained from a total time derivative ($\frac{\mathrm{d}f}{\mathrm{d}t}$), your Lagrangian is equivalent to the Lagrangian $L=0$.
So it is not surprising that your Lagrangian does not give unique equations of motions, in fact it does not give any equations of motions at all.
This also happens when there is gauge symmetry in a physical system. Then the equations of motions don't uniquely determine the evolution of all variables used to describe the system. In your system the only degree of freedom is a gauge symmetry, so there is no physical content in your Lagrangian.
As you may know, the geodesic equation, your equation (1), is not obtained as the Euler-Lagrange equations of the curve-length functional (2), but rather as the Euler-Lagrange equations of the energy functional
$$E = \frac12\int d\lambda\, g_{\mu\nu} \dot{x}^\mu\dot{x}^\nu.$$
I'm writing $\lambda$ rather than $\tau$ to avoid the suggestion that this has to be the proper time.
It is not very hard to show that extremals of $E$ are extremals of $L$, but the converse doesn't hold, in fact, length extremizing curves are extrema of $E$ if and only if they are true geodesics, i.e. affinely parameterized.
So, your equation (1) are the Euler-Langrange equations of $E$, whose solutions already are affinely parameterized. Adding (3) to it, the only additional requirement is for the curve to be timelike.
All three classes of geodesics, timelike, spacelike and lightlike, have affine parameterizations. For timelike geodesics proper time can be taken as an affine parameter, for spacelike geodesics proper length can be taken, and for lightlike curves no affine parameter has a special meaning.
Best Answer
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!
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.