What form for Transfer Functions to use for filters and Bode plots?
The first version in in the classical form $$H(s)=N(s)/D(s)$$ where both are written as $$(a_1+s)(a_2+s)\cdots (a_n+s)$$.

The second form explicitely displays the pole and zero frequency and the DC gain: $$A_0 \frac{1}{(1+s/\omega_0)(1+s/\omega_1)}$$ etc.

As you said, both forms are identical and independent of filter. They are just two (equally valid) representations and equally correct.

In my opinion, the first one is more frequently used in control theory whereas the second more in circuits.

I personally prefer the second because you immediately see the poles and zeros and the DC gain.

If you go to higher order systems, you realize that for the first representation it is not as straight forward to "see" the DC gain at first glance. You have to multiply them out:

$$H(s)=\frac{(b_1+s)(b_2+s)}{(a_1+s)(a_2+s)} = \frac{b_1 b_2}{a_1 a_2}  \frac{(1+\frac{s}{b_1})(1+\frac{s}{b_2})}{(1+\frac{s}{a_1})(1+\frac{s}{a_2})} $$

In your specific example, there is no DC gain since it has inductive character. However, L/R1 is the "differentiator gain".

In my opinion, the following example makes it more clear: Suppose you have an ideal integrator followed by a doublet:

$$
H(s) = \frac{1}{s} K \frac{s + z_1}{s + p_1}
$$

Assume the doublet occurs after the integration frequency (e.g. a parasitic pole and a zero due to finite output resistance of an opamp). Can you read off from this immediately how much "V per s" the integrator integrates from the formula above?

However, if I factor z1 and p1:

$$
H(s) = \frac{1}{s} \frac{K z_1}{p_1} \frac{1 + \frac{s}{z_1}}{1 + \frac{s}{p_1}}
$$

Note that for $$s \gg \{z_1,p_1\}$$:

$$
\frac{1 + \frac{s}{z_1}}{1 + \frac{s}{p_1}} \approx 1
$$

Now you can quickly see that the integration constant is K z1/p1.