How can we calculate the transfer function from this filter?
Most of you reasonning concerning the Vout and Vin is correct. To find the transfer function, you need to do the Voltage divider.

$$V_{out} = V_{in}\frac{R_{1}}{{R_{1} + \frac{1}{j\omega C_{1}}}} $$

if we manipulate the equation:

$$\frac{V_{out}}{V_{in}}= \frac{j\omega}{{j\omega +\frac{1}{C_{1}R_{1}}}} $$

the cuf off frequency in this problem is:

$$\omega_{c}= \frac{1}{CR} $$

so the equation become:

$$H(j\omega) =\frac{V_{out}}{V_{in}}= \frac{j\omega}{j\omega + \omega_{c}} $$

So there you have the general transfer equation. The op-amp is just a follower with unity Gain.

You can rearrange the equation to fit your format

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