Christoffel symbol for Schwarzschild metric I know that the christoffel (second kind) can be defined like this:
$$\Gamma^m_{ij} = \frac{1}{2} g^{mk}(\frac{\partial g_{ki}}{\partial U^j}+\frac{\partial g_{jk}}{\partial U^i}-\frac{\partial g_{ij}}{\partial U^k})$$
but I don't know how $U^i$ is defined (specifically for the Schwarzschild metric.
 A: The Schwarzschild metric is, in $-+++$ sign convention and units of $c = 1$ is
$$\mathrm{d}s^2 = -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \frac{\mathrm{d}r^2}{1-\frac{2M}{r}} + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\text{.}$$
We can index the coordinates arbitrarily, but let's take them in the typical order: $(U^0,U^1,U^2,U^3) = (t,r,\theta,\phi)$. In the metric, terms like $\mathrm{d}t^2$ are shorthand for the tensor product $\mathrm{d}t\otimes\mathrm{d}t$ and cross-terms like $\mathrm{d}t\,\mathrm{d}r$ for $\frac{1}{2}\left(\mathrm{d}t\otimes\mathrm{d}r+\mathrm{d}r\otimes\mathrm{d}t\right)$, since the metric must be symmetric. But we don't have any cross-terms, so the covariant metric components form a diagonal matrix:
$$g_{ij} = \begin{bmatrix}-\left(1-\frac{2M}{r}\right) &0 &0 &0\\
0&\left(1-\frac{2M}{r}\right)^{-1} &0&0\\0&0&r^2&0\\0&0&0&r^2\sin^2\theta\\
\end{bmatrix}\text{,}$$
while the contravariant matrix components $g^{ij}$ form the matrix inverse of the above, which in the case of diagonality just simplifies to $g^{ij} = 1/g_{ij}$ if $i = j$ and $g^{ij} = 0$ otherwise.
Write $_{,n}$ for the partial derivative with respect to $U^n$. Then the connection coefficients / Christoffel symbols
$$\Gamma^m_{ij} = \frac{1}{2}g^{mk}\left[g_{ki,j} + g_{kj,i} - g_{ij,k}\right]$$
simplify in the diagonal case to just
$$\Gamma^m_{ij} = \frac{1}{2}g^{mm}\left[g_{mi,j} + g_{mj,i} - g_{ij,m}\right]\text{,}$$
since terms with $m\neq k$ have $g^{mk} = 0$, and in the Schwarzschild case the metric components are independent of either $t$ or $\phi$, so:
$$\begin{eqnarray*}
g_{ij,0} = \frac{\partial g_{ij}}{\partial t} = 0&\quad\text{and}\quad&
g_{ij,3} = \frac{\partial g_{ij}}{\partial \phi} = 0\text{.}
\end{eqnarray*}$$
Thus for the Schwarszchild case, any off-diagonal term is zero and all partials by  $t$ or $\phi$ are also zero. For example,
$$\begin{eqnarray*}
\Gamma^\phi_{\theta\phi} = \Gamma^3_{23} &= \frac{1}{2}\underbrace{\left(r^2\sin^2\theta\right)^{-1}}_{g^{33}}\left(\underbrace{g_{32,3}}_0 + g_{33,2} - \underbrace{g_{23,3}}_0\right)\\
&= \frac{1}{2}\left(r^2\sin^2\theta\right)^{-1}\underbrace{\left(r^2\cdot2\sin\theta\cos\theta\right)}_{\partial_\theta(r^2\sin^2\theta)}= \cot\theta\text{.}\end{eqnarray*}$$
$$\Gamma^\phi_{r\phi} = \Gamma^3_{13} = \ldots = r^{-1}\text{,}$$
and other than terms required by symmetry of lower indices ($\Gamma^\phi_{\phi\theta} = \Gamma^\phi_{\theta\phi}$, &c.), those are the only nonzero $\Gamma^\phi_{ij}$. You should be able to find the rest yourself. 
P.S. There are alternative ways of finding the connection coefficients other than the formula you're using, usually from a Lagrangian, but for diagonal metrics this direct approach isn't that bad at all.
