Say I have a hollow cylinder and I wanted to strap it down to the bed of a truck. I would tension the strap on one end, and it would exert a force on the cylinder. My intuition tells me that the strap would crush the hollow cylinder down toward the truck bed, but when I think about it, there are inward forces perpendicular to the truck bed caused by the straps on the cylinder as well. Is this correct thinking, or are all of the forces only vertical?
You have successfully discovered that the kinetic energy depends on the reference frame.
That is actually true.  What is amazing, however, is that while the value of the energy is frame DEpendent, once you've chosen a reference frame, the law of conservation of energy itself is NOT reference frame-dependent -- every reference frame will observe a constant energy, even if the exact number they measure is different.  So, when you balance your conservation of energy equation in the two frames, you'll find different numbers for the total energy, but you will also see that the energy before and after an elastic collision will be that same number.
So, let's derive the conservation of energy in two reference frames.  I'm going to model an elastic collision between two particles.  In the first reference frame, I am going to assume that the second particle is stationary, and we have:
$$\begin{align}
\frac{1}{2}m_{1}v_{i}^{2} + \frac{1}{2}m_{2}0^{2} &= \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}\\
m_{1}v_{i}^2 &= m_{1}v_{1}^{2} + m_{2}v_{2}^{2}
\end{align}$$
to save myself time and energy, I'm going to call $\frac{m_{2}}{m_{1}} = R$, and we have:
$$v_{i}^{2} = v_{1}^{2} + Rv_{2}^{2}$$
Now, what happens if we shift to a different reference frame, moving to the right with speed v?  This is essentially the same thing as subtracting $v$ from all of these terms.  We thus have:
$$\begin{align}
(v_{i}-v)^{2} + R(-v)^{2} &= (v_{1}-v)^{2} + R(v_{2}-v)^{2}\\
v_{i}^{2} -2v_{i}v + v^{2} + Rv^{2} &= v_{1}^{2} - 2 vv_{1} + v^{2} + Rv_{2}^{2}-2Rv_{2}v + Rv^{2}\\
v_{i}^{2} -2v_{i}v &= v_{1}^{2}- 2vv_{1} + Rv_{2}^{2}-2Rv_{2}v\\
v_{i}^{2} &= v_{1}^{2} + Rv_{2}^{2} + 2v(v_{i} - v_{1} - R v_{2})
\end{align}$$
So, what gives?  It looks like the first equation, except we have this extra $2v(v_{i} - v_{1} - R v_{2})$ term?  Well, remember that momentum has to be conserved too.  In our first frame, we have the conservation of momentum equation (remember that the second particle has initial velocity zero:
$$\begin{align}
m_{1}v_{i} + m_{2}(0) &= m_{1}v_{1} + m_{2}v_{2}\\
v_{i} &= v_{1} + Rv_{2}\\
v_{i} - v_{1} - Rv_{2} &=0
\end{align}$$
And there you go!  If momentum is conserved in our first frame, then apparently energy is conserved in all frames!
