1.) An extra bias voltage is necessary because you are working with single supply only. However, I would suggest to use a split supply (if available) without such external biasing.
2.) For lowpass stages, the resistor between the first node and ground can always be set to infinity (can be removed). This removal is not necessary but this resistor complicates the dimensioning.
3.) The Sallen-Key topology is very sensitive to tolerances of the gain setting resistors. Therefore, it is recommended to use a design strategy based on unity gain values or gain of two (two equal resistors in the feedback path, any values). For this purpose, there are several filter design programs available (online or downloadable).
4.) For a 4-pole filter (as shown in the figure) you need two stages with DIFFERENT pole locations (that means: no identical stages). The parts values, of course, depend on the desired cut-off frequency and the selected approximatioin (Butterworth, Chebyshev,...). Use filter design programs for finding the values.
5.) As it seems, the online program (OKAWA, your link) works for a second-order filter only. Because you need two different stages, you need two different pole frequencies with different Q values. If required, I can give you the values (based on the specification as mentioned under 4.).
UPDATE: As mentioned under 1.) the dc bias of +5V is necessary because the negative supply pin of the opamps is at ground. It is best to use a symmetrical +/-12 volts supply for the opamps. In this case, of course the +5V are not required. Connect this resistor simply to ground. This resistor is necessary to allow a small dc bias current for the opamp input (the value may be larger if you need a larger input resistance of the whole circuit.)
Your formulas do not take into account that the impedance of a capacitor is complex !
What you use is:
$$Zc(f) = 1 / 2 \pi f$$
but what you should use is:
$$Zc(s) = 1 / s$$
where $$s = 2 \pi f j $$ where j makes it imaginary remember: $$j^2 = -1$$
Have a look at this page on Wikipedia
where they show how to determine the transfer function in the s domain.
Best Answer
I am not using classical node-mesh analysis here but rather the fast analytical circuits techniques (FACTs). I am not saying they are the best path here but I could derive the transfer function of this second-order filter quite quickly with the help of a few individual sketches I could fix in case of problems. The generalized transfer function of a 2nd-order filter is defined as
\$H(s)=\frac{H_0+s(H^2\tau_2+H^3\tau_3)+s^2H^{32}\tau_3\tau_{32}}{1+s(\tau_2+\tau_3)+s^2(\tau_3\tau_{32})}\$
The time constants in the denominators are found by setting the excitation to 0 (reduce \$V_{in}\$ to 0 V and replace it by a short circuit) and "looking" at the individual resistance driving each capacitor. The resistance multiplied by the capacitor it drives forms the time constants \$\tau\$ we want to determine the denominator \$D(s)\$. Then, we look for the gain of this circuit in dc, when \$s=0\$ and it gives us \$H_0\$. Finally, we determine the gain of this filter when some of its capacitors are set in their hi-frequency state. For instance, if I replace \$C_3\$ by a short circuit (its hi-frequency state), the gain \$H^3=0\$ and you continue with the rest of the gains to form \$N(s)\$. The sketches appear below:
When you have determined all these time constants (you obviously need KVL and KCL here), you gather all the elements in Mathcad to obtain the final expression that I put in a low-entropy form, making the band-pass gain clearly appearing.
and it you plot the whole thing, you have:
The FACTs are an excellent way of deriving transfer functions in a swift and efficient manner. I have gone through individual sketches to determine the \$a_i\$ and \$b_i\$ of \$N(s)\$ and \$D(s)\$ meaning that if I observe a deviation at the end between simulation and the Mathcad plot, I can fix the guilty coefficient immediately, something I could not do with the classical approach. If you want to know more about FACTs, have a look at the seminar taught at APEC 2016
http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf
but also at the numerous transfer functions derived in the book
http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf