Histogram Intersection with two different bin sizes?
I don't see any other way than to adapt one of the histograms such that both histograms have the same number of bins before performing histogram intersection. This will always involve some guesswork since the histograms themselves don't give you enough information to do this accurately. I would adapt the histogram with the larger number of bins to the number of bins of the other histogram. This is a simple example:

h1 = [12 17]
h2 = [9 10 11]


So in this case you want to change h2 such that it has two bins like h1. Since the interval boundary between the two intervals of h1 is right in the middle of bin 2 of h2, it is reasonable to divide bin 2 of h2 evenly between the two other bins. So you get

h2_new = [14 16]


If the new boundary falls somewhere else inside a bin interval just divide the interval by the respective percentage. This step of course involves guesswork and it would be much better if you could avoid it by making sure that both histograms are computed using the same number of bins.

As a final note, usually histogram intersection is normalized such that its maximum value is $1$:

$$K(a,b)=\frac{\sum_{i=1}^n\min (a_i,b_i)}{\min\left(\sum_i a_i,\sum_i b_i\right)}$$