compact support intuition needed

Why do we require that $\varphi(x) = 0$ for $x$ large enough ? Because it allows us to integrate by parts without fear : $$\int_{-\infty}^\infty T(x) \varphi'(x)dx = \lim_{x \to \infty} T(x)\varphi(x)-T(-x)\varphi(-x)-\int_{-\infty}^\infty T'(x) \varphi(x)dx$$ Here $\varphi \in C^\infty_c$ and $T$ is a distribution, so $T(x)\varphi(x)$ doesn't make sense, but if you assume that $\varphi(x) = 0$ for $|x| > M$ then clearly $\lim_{x \to \infty} T(x)\varphi(x)-T(-x)\varphi(-x) = 0$ and it makes sense to write $$\langle T,\varphi' \rangle =\int_{-\infty}^\infty T(x) \varphi'(x)dx = -\int_{-\infty}^\infty T'(x) \varphi(x)dx=-\langle T',\varphi \rangle \tag{1}$$ which is exactly what we need for defining $\delta'$ the derivative of the Dirac delta and the derivatives of distributions in general.

Now the big question : how do you prove that $(1)$ makes sense (that it doesn't lead to some contradictions) ? Well you can take it as a definition, so nothing to prove,

or you can show that when defining the distributions as linear operators $C^\infty_c \to \mathbb{R}$ continuous for the test function space topology, then the differentiation operator $\langle T,.\rangle \mapsto \langle T',.\rangle$ is continuous in the sense of distributions.

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See also the Schwartz space, where we replace the compact support property by a decay $o(x^{-k})$ at $\infty$, discarding the distributions with a too large grow rate and keeping the so-called tempered distributions.

The functions with compact support are those that are zero outside of a compact set. This is quite useful in distributions as it tells us that the function dosent grow indefinitely. It is useful also in the theory of Differential Equations, Functional Analysis, Toplogy. Some examples include Bump Function. Also refer Support for different types of support.