The rate equation for this type of reaction is:

$$ Rate = k[NO]^x [H_2]^y /$$

As you had written a good way to determine the reaction order of NO is to look at two reactions done with the same concentration of H_{2} but different concentrations of NO. The ratio of the reaction rates is give by:

$$ \frac{rate_2}{rate_1} = \frac{k [NO]_2^x[H_2]_2^y}{k [NO]_1^x [H_2]_1^y /$$

Since the concentration of H_{2} is the same for both reactions 1 and 2 and since the rate constant is a constant, we can cancel those two factors and come up with a simplified expression:

$$ \frac{rate_2}{rate_1} = \left( \frac{[NO]_2}{[NO]_1} \right)^x /$$

Now, as you note, rate_{2}/rate_{1} = 4 while [NO]_{2}/[NO]_{1} = 2. Since *doubling* the concentration of [NO] leads to a *quadrupling* of the rate, the reaction must be second order with respect to NO. If doubling the concentration of NO had led to only a 2-fold increase in the rate of reaction, then the reaction would be first order with respect to NO. Similarly, if doubling the concentration of NO had lead to an eight fold increase in the rate of reaction, the reaction would be third-order with respect to NO.