1.why is renormalization even necessary?
It is because, experiments force us to do that. You suppose to get finite values i.e, for charge or for mass of a particle.
2.if we always get a single finite result, why do we care that there are infinities in the equations at the level of amplitudes?
Actually we don't. Your integrals in the loop corrections, would be infinite and the way you get rid of them is counter-terms. You cancel the infinite part of the loop corrections with counter-terms and then finite parts still remains. In QED. after regularization and cancellation of infinities, you will get something like $M=iC\lambda^2 log(\frac{\mu^2}{s})$. M amplitude, C, numerical numbers, s, center of mass energy and $\mu^2$ is what we put cut off for energy scale.But do not forget, this is what we got after regularization and cancel the infinite parts with the counter-terms. If one wants to calculate the amplitude, need to know cut off scale. For an experimentalist, $\mu$ makes perfect sense that any given theory has a finite domain of validity. 651
But what is the precise value? If someone wants to change $\mu$ what happens? M is suppose to be finite and should not depend on value $\mu$. On the other hand, we also do not know the value of $\lambda$. What is $\lambda$? It is basically a coupling constant, here not a physical parameter. If someone wants to change $\mu$, one just shift the $\lambda$ in such a way that M does not change!So $\lambda$ is secretly a function of $\mu$. This gives rise to RGE. But this is not the topic here. We have still no information about cut off scale. How can we get rid of it?
Here $\lambda$, as I said in the first paragraph, not a physical value. And the expression above, depends on an arbitrary scale $\mu$.In observables, one need to define $\lambda$ as a physical parameter(observable). I mean, If we renormalize the theory at some scale i.e, $s_0$ then the cut off scale vanishes and the lagrangian comes out as, $$ L-L_0=iC\lambda^2 log(\frac{s_0}{s}) $$ Then you can be sure that when the scattering amplitude is expressed in terms of physical coupling constant $\lambda_p$, the cut off disappear completely! And all the values remain physical and observable. 651