I'd like to expand on Hans Lundmark's answer because the question keeps recurring.
Let $q^1,\ldots,q^n$ denote the generalized coordinates. Introduce additional velocity variables $\dot{q}^1,\ldots,\dot{q}^n$. If the dot accent is already reserved for other things in your theory, use another accent to avoid confusion.
First, treat all $q^i$ and $\dot{q}^i$ as pairwise independent variables and define, using Einstein summation convention, $L(q^1,\ldots,q^n;\dot{q}^1,\ldots,\dot{q}^n) = \frac{1}{2} g_{ij}(q^1,\ldots,q^n)\,\dot{q}^i \dot{q}^j\tag{1}$ To avoid confusion later on, I have explicitly indicated the formal dependencies of the expressions for $g_{ij}$ and $L$. Note that $L$ can immediately be written down when given the first fundamental form.
Now consider a twice differentiable curve, which makes the $q^i$ functions of some independent new parameter $\tau$, and set $\dot{q}^i = \frac{\mathrm{d}q^i}{\mathrm{d}\tau}$.
Proposition. On the curve parameterized with $\tau$ we have $g^{kh}\left(\frac{\mathrm{d}}{\mathrm{d}\tau} \left(\frac{\partial L}{\partial\dot{q}^h}\right) - \frac{\partial L}{\partial q^h}\right) = \ddot{q}^k + \Gamma^k_{\ ij} \dot{q}^i \dot{q}^j \tag{2}$ You might recognize the right-hand side as the expression constrained by geodesics, and you might recognize the expression wrapped around $L$ in the left-hand side as the form of Euler-Lagrange differential equations, which correspond to variational problems with the Lagrangian $L$. Therefore $(2)$ hints at a variational foundation for the geodesics equation. Such interpretations make $(2)$ easier to memorize or cross-link with other knowledge, but all we actually need is the identity $(2)$. The following observations should provide enough details to prove $(2)$.
We can rewrite $(2)$ to $\frac{\mathrm{d}}{\mathrm{d}\tau} \left(\frac{\partial L}{\partial\dot{q}^h}\right) - \frac{\partial L}{\partial q^h} = g_{hk} \ddot{q}^k + \Gamma_{hij} \dot{q}^i \dot{q}^j \tag{3}$ The idea now is, for every $h\in\{1,\ldots,n\}$, to take the left-hand side of $(3)$, plug in expressions for the metric in $L$, and rewrite the thing so that it matches the format of the right-hand side, where dotted variables occur only in the places shown. Then you can read off the Christoffel symbols of the first kind, $\Gamma_{h**}$, from the coefficients of the velocity products.
To obtain $(2)$ and $\Gamma^k_{\ **}$ is then just a matter of multiplication with the inverse metric coefficients matrix $((g^{kh}))$, or equivalently, taking the right-hand side expressions of $(3)$ obtained for $h\in\{1,\ldots,n\}$ and then, for each $k\in\{1,\ldots,n\}$, finding a linear combination whose only second derivative with respect to $\tau$ is $\ddot{q}^k$, with coefficient $1$. This is the form given by $(2)$, so the coefficients of the velocity products are then the Christoffel symbols of the second kind, $\Gamma^k_{\ **}$.
Remember, while doing the partial derivatives of $L$, treat the $q^i$ and the $\dot{q}^i$ as independent formal variables. You will do that with more concrete symbol meanings and metric expressions, but in this moderately abstract setting, you can already refine $\begin{align} \frac{\partial L}{\partial\dot{q}^h} &= g_{hj}\dot{q}^j\tag{4} \\\frac{\partial L}{\partial q^h} &= \frac{1}{2}\frac{\partial g_{ij}}{\partial q^h}\dot{q}^i \dot{q}^j\tag{5} \end{align}$ However, when doing the $\frac{\mathrm{d}}{\mathrm{d}\tau}$ outside of $L$, stick to the curve and apply the chain rule accordingly: $\frac{\mathrm{d}g_{hj}}{\mathrm{d}\tau} = \frac{\partial g_{hj}}{\partial q^i}\,\dot{q}^i\tag{6}$ Now $(4)$, $(5)$, $(6)$ and the Levi-Civita formula $\Gamma_{hij} = \frac{1}{2}\left( \frac{\partial g_{hj}}{\partial q^i} + \frac{\partial g_{ih}}{\partial q^j} - \frac{\partial g_{ij}}{\partial q^h} \right)$ can be used to prove $(3)$ and thereby $(2)$. But I will focus on how to apply that proposition.
Example: Spherical coordinates with radius $r$, longitude $\phi$, latitude $\theta$, with $\theta=\frac{\pi}{2}$ at the equator. At index positions, I will write coordinate names instead of digits. The first fundamental form is $\mathrm{d}s^2 = \mathrm{d}r^2 + (r^2\sin^2\theta)\,\mathrm{d}\phi^2 + r^2\,\mathrm{d}\theta^2$ Accordingly, the Lagrangian $L$ is $L = \frac{1}{2}\left(\dot{r}^2 + (r^2\sin^2\theta)\,\dot{\phi}^2 + r^2\,\dot{\theta}^2\right)$ We now treat $r,\phi,\theta,\dot{r},\dot{\phi},\dot{\theta}$ as independent variables and get $\begin{align} \frac{\partial L}{\partial\dot{r}} &= \dot{r} &\frac{\partial L}{\partial r} &= (r\sin^2\theta)\,\dot{\phi}^2 + r\,\dot{\theta}^2 \\\frac{\partial L}{\partial\dot{\phi}} &= (r^2\sin^2\theta)\,\dot{\phi} &\frac{\partial L}{\partial\phi} &= 0 \\\frac{\partial L}{\partial\dot{\theta}} &= r^2\,\dot{\theta} &\frac{\partial L}{\partial\theta} &= (r^2\sin\theta\cos\theta)\,\dot{\phi}^2 \end{align}$ Now we give up the independence, consider some curve paramterized by $\tau$ and obtain $\begin{align} \frac{\mathrm{d}}{\mathrm{d}\tau} \frac{\partial L}{\partial\dot{r}} &= \ddot{r} \\\frac{\mathrm{d}}{\mathrm{d}\tau} \frac{\partial L}{\partial\dot{\phi}} &= (r^2\sin^2\theta)\,\ddot{\phi} + 2\,(r\sin^2\theta)\,\dot{r}\,\dot{\phi} + 2\,(r^2\sin\theta\cos\theta)\,\dot{\phi}\,\dot{\theta} \\\frac{\mathrm{d}}{\mathrm{d}\tau} \frac{\partial L}{\partial\dot{\theta}} &= r^2\,\ddot{\theta} + 2\,r\,\dot{r}\,\dot{\theta} \end{align}$ And so $\begin{align} \frac{\mathrm{d}}{\mathrm{d}\tau} \left(\frac{\partial L}{\partial\dot{r}}\right) - \frac{\partial L}{\partial r} &= \underbrace{1}_{g_{rr}}\,\ddot{r} + \underbrace{(-r\sin^2\theta)}_{\Gamma_{r\phi\phi}}\,\dot{\phi}^2 + \underbrace{(-r)}_{\Gamma_{r\theta\theta}}\,\dot{\theta}^2 \\\frac{\mathrm{d}}{\mathrm{d}\tau} \left(\frac{\partial L}{\partial\dot{\phi}}\right) - \frac{\partial L}{\partial\phi} &= \underbrace{(r^2\sin^2\theta)}_{g_{\phi\phi}}\,\ddot{\phi} + 2\,\underbrace{(r\sin^2\theta)}_{\Gamma_{\phi r\phi} = \Gamma_{\phi\phi r}}\,\dot{r}\,\dot{\phi} + 2\,\underbrace{(r^2\sin\theta\cos\theta)}_{\Gamma_{\phi\phi\theta} = \Gamma_{\phi\theta\phi}}\,\dot{\phi}\,\dot{\theta} \\\frac{\mathrm{d}}{\mathrm{d}\tau} \left(\frac{\partial L}{\partial\dot{\theta}}\right) - \frac{\partial L}{\partial\theta} &= \underbrace{r^2}_{g_{\theta\theta}}\,\ddot{\theta} + 2\,\underbrace{r}_{\Gamma_{\theta r\theta} = \Gamma_{\theta\theta r}}\,\dot{r}\,\dot{\theta} + \underbrace{(-r^2\sin\theta\cos\theta)}_{\Gamma_{\theta\phi\phi}} \,\dot{\phi}^2 \end{align}$ All other Christoffel symbols of the first kind are zero. If we had a non-diagonal metric, some right-hand side expressions would have several second derivatives, each accompanied by a corresponding metric coefficient.
To obtain the Christoffel symbols of the second kind, find linear combinations of the above right-hand side expressions that leave only one second derivative, with coefficient $1$. Here this is easy because the metric is already in diagonal form. Therefore $\begin{align} g^{rh}\left(\frac{\mathrm{d}}{\mathrm{d}\tau} \left(\frac{\partial L}{\partial\dot{q}^h}\right) - \frac{\partial L}{\partial q^h}\right) &= \ddot{r} + \underbrace{(-r\sin^2\theta)}_{\Gamma^r_{\ \phi\phi}}\,\dot{\phi}^2 + \underbrace{(-r)}_{\Gamma^r_{\ \theta\theta}}\,\dot{\theta}^2 \\g^{\phi h}\left(\frac{\mathrm{d}}{\mathrm{d}\tau} \left(\frac{\partial L}{\partial\dot{q}^h}\right) - \frac{\partial L}{\partial q^h}\right) &= \ddot{\phi} + 2\,\underbrace{\left(\frac{1}{r}\right)}_{\Gamma^\phi_{\ r\phi} = \Gamma^\phi_{\ \phi r}}\,\dot{r}\,\dot{\phi} + 2\,\underbrace{(\cot\theta)}_{\Gamma^\phi_{\ \phi\theta} = \Gamma^\phi_{\ \theta\phi}}\,\dot{\phi}\,\dot{\theta} \\g^{\theta h}\left(\frac{\mathrm{d}}{\mathrm{d}\tau} \left(\frac{\partial L}{\partial\dot{q}^h}\right) - \frac{\partial L}{\partial q^h}\right) &= \ddot{\theta} + 2\,\underbrace{\left(\frac{1}{r}\right)}_{\Gamma^\theta_{\ r\theta} = \Gamma^\theta_{\ \theta r}}\,\dot{r}\,\dot{\theta} + \underbrace{(-\sin\theta\cos\theta)}_{\Gamma^\theta_{\ \phi\phi}} \, \dot{\phi}^2 \end{align}$ All other Christoffel symbols of the second kind are zero.