Intuitively, why do attempts to delay hitting a black hole singularity cause you to reach it faster?
Actually, it turns out to be incorrect that the optimal strategy is to free fall. There is an optimal strategy for firing your rocket engine which maximizes your proper time from the event horizon to the singularity, and extends it beyond the proper time of a free falling observer.
Here is a paper that discusses the issue and describes strategies for maximizing the proper time to the singularity:
https://arxiv.org/abs/0705.1029v2
Edit: a TL;DR summary of the paper. An infalling rocket can maximize the proper time to the singularity by first making a burn to match the trajectory of a free falling object which started at rest at the horizon. Once the rocket has matched that specific trajectory, then it should turn off engines.
My (very limited) intuition for this is that once you cross the event horizon, the singularity is not so much a distant point in space as it is a moment in future time.
In other words, within the event horizon you're firing your rockets not to avoid some point $(x,y,z)$, but rather to avoid next Thursday. From here, I use my intuition about time dilation and the fact that geodesics are trajectories of maximum proper time.
I'm by no means a GR expert so if this picture is wrong, corrections are more than welcome :)
Here's a partial answer, although it's still pretty formal. First define $$E := -\left( \frac{2GM}{r} - 1 \right) \frac{dt}{d\tau}, \qquad L := r^2 \frac{d\phi}{d\tau}$$ in the usual Schwarzschild coordinates. If you're free falling, then $E$ and $L$ are constant over your trajectory, but if you can fire your engines then they can change. We can expand out the normalization condition $U \cdot U = -1$ to $$\frac{1}{2} \left( \frac{dr}{d\tau} \right)^2 - \left(\frac{2GM}{r} - 1 \right) \left(1 + \frac{L^2}{r^2} \right) = \frac{1}{2} E^2,$$ which looks somewhat like the statement of conservation of energy (per unit mass) for a nonrelativistic particle with angular momentum $L$.
But there are two weird aspects to this equation:
When you expand out the product of the two binomials on the LHS, you get a weird term $-2GM L^2/r^3$ that does not appear in the nonrelativistic case. Unlike the usual centrifugal angular momentum barrier, this is a centripetal angular momentum fictitious "force" that actually sucks the particle inward at small radii. This means that angular momentum is actually your enemy, not your friend, for avoiding the singularity - so you don't want to accelerate in a way that increases its magnitude.
The effective total energy $\mathcal{E}$ isn't the physical mechanical energy $E$, but instead $\frac{1}{2} E^2$. In the standard nonrelativistic case, firing your engines to slow your infall decreases your total mechanical energy and helps you delay getting close to the center. (This may seem counterintuitive at first, because we associate highly negative energies with tightly bound orbits and positive energies with unbound orbits, so you might think you would want to increase your energy. But for the purpose of delaying getting close to the center, you actually want to brake and make your energy more negative, at the expense of trapping yourself deeper in the gravity well overall and spending more time near the center once you finally do get there.) But in the Schwarzschild case, $\mathcal{E} = \frac{1}{2} E^2$ means that your effective energy actually depends non-monotonically on your physical energy: if your physical energy $E$ is negative, then making it even more negative actually increases your effective energy $\mathcal{E}$. This means that minimizing your effective energy requires keeping your physical energy at $E = 0$, which indeed corresponds to the optimal geodesic which begins at rest infinitesimally outside the horizon. Any attempt to brake further will overshoot $E = 0$ and send $E$ negative, which will actually increase your effective energy and hurt you.