my Butterworth lowpass formulas do not agree with Fisher webpage
The problem is in the way you apply the bilinear transform. You have to use the appropriate (pre-)warping of the frequencies. Since the bilinear transform warps the frequency axis, you have to make sure that the corner frequency of the discrete-time filter is correct. One way to do that is as follows. The bilinear transform is defined as

$$s=k\frac{z-1}{z+1}\tag{1}$$

with some constant $k$ yet to be defined. If we denote the analog frequency by $\Omega$, and the normalized discrete-time frequency by $\omega$, Eq. (1) becomes for $s=j\Omega$ and $z=e^{j\omega}$

$$j\Omega=k\frac{e^{j\omega}-1}{e^{j\omega}+1}=k\frac{e^{j\omega/2}-e^{-j\omega/2}}{e^{j\omega/2}+e^{-j\omega/2}}=jk\tan(\omega/2)\tag{2}$$

Eq. (2) describes the frequency warping caused by the bilinear transform. If we use an analog lowpass filter with a normalized corner frequency $\Omega_0=1$, we must choose the constant $k$ such that for the desired discrete-time corner frequency $\omega_0$ the term on the right-hand side of Eq. (2) becomes $1$:

$$k=\frac{1}{\tan(\omega_0/2)}\tag{3}$$

where $\omega_0$ is the desired corner frequency of the digital filter. Eq. (3) and Eq. (1) define the appropriately normalized bilinear transform that you must use.

So for your example, the normalized first-order analog Butterworth lowpass transfer function is given by

$$H(s)=\frac{1}{1+s}\tag{4}$$

Applying the bilinear transform gives

$$H(z)=\frac{1}{1+k\frac{z-1}{z+1}}=\frac{z+1}{z(1+k)+1-k}=\frac{1}{1+k}\cdot\frac{1+z^{-1}}{1+\frac{1-k}{1+k}z^{-1}}\tag{5}$$

Ignoring the gain term $1/(1+k)$, the corresponding difference equation is

$$y[n]=x[n]+x[n-1]-\frac{1-k}{1+k}y[n-1]\tag{6}$$

With a desired corner frequency $\omega_0=2\pi/30$ you get from (3) $k=9.5144$ and $(1-k)/(1+k)=-0.80978$, just like on Fisher's website.