Since the Levi-Civita connection is unique, the easiest way to determine it is often educated guesswork. We illustrate the technique using polar coordinates.

Our orthonormal basis of 1-forms is $\{dr,r\,d\phi\}$. Writing out the torsion-free condition, we have \begin{align} d(dr) + \omega^r{}_r\wedge dr + \omega^r{}_\phi\wedge r\,d\phi &= 0 \\ d(r\,d\phi) + \omega^\phi{}_r\wedge dr + \omega^\phi{}_\phi\wedge r\,d\phi &= 0 \end{align} But metric compatibility tells us that \begin{align} \omega^r{}_r = & 0 = \omega^\phi{}_\phi \\ \omega^\phi{}_r &= -\omega^r{}_\phi \end{align} so that \begin{align} \omega^r{}_\phi\wedge r\,d\phi &= 0 \\ dr\wedge d\phi - \omega^r{}_\phi\wedge dr &= 0 \end{align} In order to solve the second of these equations, $\omega^r{}_\phi$ must contain a term $-d\phi$. Guessing \begin{equation} \omega^r{}_\phi = -d\phi \end{equation} not only solves the second equation, but also the first, so we are done. A quick comparison with \begin{equation} \omega_{r\phi} = \rhat\cdot d\phat = -d\phi \end{equation} shows that our answer agrees with our previous computations; there are no signs to worry about when “lowering an index” in this Euclidean example.

Yes, we could have solved the first equation explicitly, thus showing that there is no $dr$ term in $\omega^r{}_\phi$, and in this particular example there's not much difference between these two solution strategies. But in more complicated examples, it is often most efficient to simply assume that certain components are zero; if it works, you're done.