Intuition is just pattern recognition. This comes with doing many problems that force you to think and keep your brain completely engaged with the material. I remember when I first started physics, I thought it was impossible to keep track of all of the types of problems. There were balls rolling down inclined planes and masses on pulleys and people pushing blocks up hills with friction. Then there was a certain point when a lightbulb went off and it made sense: these are all the same problem. All you do is figure out where all the forces are pointing and how strong they are, then add them up to find the net force, and you've got the equation of motion in hand. Eventually, it becomes second nature, so I can glance at a problem in mechanics and pretty quickly have an intuition for what should happen.
The same thing happens in more advanced physics. You start to recognize classes of physical situations and types of symmetry. Let me give you a a more advanced example that was a similar lightbulb for me - it's a bit lengthy, and you don't need to understand every step. The important part is the conclusion. Try evaluating this integral:
$\int_{-\infty}^{\infty}e^{ip(x-x_0)/\hbar}e^{-|p|/p_0}dp$
This looks horrible, doesn't it? And yes, if you just charge in and try to integrate it, you'll wind up doing a long series of painful integrations by parts. But there are a bunch of things here that a practiced physics student will recognize that make it easy.
Since $e^{ip(x-x_0)/\hbar}=\cos(p(x-x_0)/\hbar)+i\sin(p(x-x_0)/\hbar)$ (that's clever trick no. 1), this is actually an integral of a cosine plus a sine, multiplied by a decaying exponential, over $(-\infty,\infty)$. But sine is an odd function, so it evaluates to zero over $(-\infty,\infty)$ (clever trick no. 2)! That means the integral is actually
$\int_{-\infty}^{\infty}\cos(p(x-x_0)/\hbar)e^{-|p|/p_0}dp$ Now, we have an even function inside the integral, so its value over $(-\infty,\infty)$ is twice its value over $(0,\infty)$ (clever trick no. 2, reused). Furthermore, $\cos(p(x-x_0)/\hbar)=Re(e^{ip(x-x_0)/\hbar})$ (going backwards with clever trick no. 1). This makes the integral
$2Re(\int_{0}^{\infty}e^{ip(x-x_0)/\hbar}e^{-p/p_0}dp)$
$2Re(\int_{0}^{\infty}e^{(i(x-x_0)/\hbar - 1/p_0)p}dp)$
$2Re[\frac{1}{i(x-x_0)/\hbar + 1/p_0}]$
$\frac{2p_0}{1+(x-x_0)^2p_0^2/\hbar^2}$ (Maybe clever trick no. 3 if you aren't very used to complex numbers?)
Here's the kicker: If you do it enough, this sort of integral starts coming intuitively and quickly. The even-and-odd tricks become obvious; the conversion back and forth between exponential and trigonometric functions becomes totally natural. An integral like this flies by at the speed of thought, and I don't need to write out all those steps - it just feels sort of obvious at each point. That's just pattern recognition from doing it so many times - or, if you like, "intuition."
If you want to build up your intuition, do many, many problems. As many as you can get your hands on. That's what it takes.