Transfer functions are expressions that represent an output quantity divided by an input quantity. You do not specify an input or output quantity here.
You also need a reference voltage. Voltages are specified as potential differences between points. Without a reference, you would not be able to say node 4 is 5 or 12 or 10000 volts; you could only say that node 4 is 5 or 10 or 1000 volts greater than node 3.
Once you find out what your input, output, reference point, and reference voltage are, you can proceed with finding the transfer function. There are a few different techniques you can use, but I like to use the mesh current method. I sketched an example of the mesh current method for your circuit below. An internet search of mesh current method will tell you how to do it. The mesh current method will give you a system of N linear equations where N is the number of current loops in your network.
You can solve this system of linear equations by hand, or you can write them in matrix form Z*I = V where Z contains all of your R's, C's, and L's; I is your current loops; and V is your voltage. You can then solve for the currents I with a matrix solver like numpy in Python or MATLAB using I = inv(Z)*V. This is a good way to check your answer or solve larger networks. The link below describes this method.
http://www.analyzemath.com/applied_mathematics/electric_circuit_1.html
Once you have solved for your mesh currents, you can find any voltage difference. In this example, the voltage across R2 is R2*(I1-I2). Now, if we know that the reference voltage is 0 volts and the reference point is node 2, then we can say that the voltage at node 3 is 0 + R2*(I1-I2).
Once you know the quantity (a voltage in this case) at your output node, you can find the transfer function by dividing by your input quantity (a voltage in this case). This will be a rational function, and the roots of the denominator are called the poles and the roots of the numerator are called the zeros. Your cutoff frequency is the pole of your transfer function. If you have multiple poles, you will have multiple cutoff frequencies if the poles are unique. This makes sense for a bandpass or notch filter. If the poles are the same, i.e. the denominator of the rational function has repeated roots, then you will only have 1 cutoff frequency but you will have more attenuation after the cutoff frequency compared to if there was only 1 pole at the cutoff frequency.
I get the same transfer function you did
\$H(s)=\frac{R2}{R1 + R2 + sC_1(R_1R_2+R_1R_3+R_2R_3)}\$

syms R1 R2 R3 C1 s v1
% write mesh current equations
Z = [-(R2+R1) R2;
R2 -(R2+R3+1/(s*C1))];
V = [-v1; 0];
I = inv(Z)*V;
% identify transfer function
tf = (I(2) * 1/(s*C1)) / v1;
% solve poles and zeros
[num,den] = numden(tf);
zeros = solve(num,s); % there are no zeros
poles = solve(den,s);
% numerical evaluation
vars = [R1 R2 R3 C1];
numVars = [100 1e3 1.24e3 1e-9];
cutoff = vpa(subs(poles(1), vars, numVars));