Electronic – Why is this LC filter where it is, and what function does it perform

I'll paraphrase what Dave Tweed says.

If you input a sinewave then there is no misshaping of the signal just an attenuation that is dependent on how high the input frequency is. If you input a square wave, the output will be misshaped because the harmonics (that comprise the square wave) will be attenuated at significantly different levels and you'll likely start to see an output that more resembles a sine wave or triangular wave.

Here's what a 1kHz square wave looks like when it is filtered with the same circuit as the OP has in his question: -

The input is 1Vp-p centred on 3V dc and as you can see the square wave is nearly triangular.

The transfer function is not \$H(\omega)\$, it is \$H(j\omega)\$ (note the \$j\$, which makes it complex):

$$H(j\omega) = \frac{1}{1+j\omega RC}$$

This is important because the transfer function captures the phase in addition to the amplitude.

The amplitude is

$$|H(j\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}}$$

by the definition of complex magnitude. This is the gain you measure from input to output and it may be all you care about, especially for a first order system. However, the phase $$\angle H(j\omega) = -\arctan(\omega RC)$$ can be quite important (e.g. for ensuring stability), especially for higher order transfer functions.

The main, direct use of the transfer function is to capture both gain and phase in one expression, but can also be used for time-domain analysis, as the transfer function is the Laplace transform of the impulse response. It is also useful for characterizing a multiple stage system, since the transfer function \$H(j\omega)\$ of a system consisting of stage \$H_1(j\omega)\$ followed by \$H_2(j\omega)\$ is simply \$H(j\omega) = H_1(j\omega)H_2(j\omega)\$ whereas in the time-domain you need to convolve the impulse responses \$h_1(t)\$ and \$h_2(t)\$ to find the overall system impulse response of \$h(t)\$.