Suppose $D \in \mathcal L(\mathcal P_3(\mathbf R), \mathcal P_2(\mathbf R))$ is the differentiation map defined by $Dp = p'$. Find a basis of $\mathcal P_3(\mathbf R)$ and a basis of $\mathbf P_2(\mathbf R)$ such that the matrix of $D$ with respect to these bases is

Consider the basis $(x^3, x^2, x, 1)$ of $\mathcal P_3(\mathbf R)$ and the basis $(3x^2, 2x, 1)$ of $\mathcal P_2(\mathbf R)$. Clearly $D(x^3) = 3x^2$ etc, meaning the matrix is as desired.

In more general terms, we picked $(x^3, x^2, x, 1)$ so that the nullspace was at the end, then picked $(Dx^3, Dx^2, Dx)$ as our output basis.

Matrix multiplication can be interpreted as writing our vectors in terms of a new basis, so picking a basis that makes that look like the identity by applying the transformation to each basis vector makes sense.

In the nice case where we're dealing with an invertible linear map $T \in \mathcal L(V,W)$ we can pick $v_1,\dots,v_n$ a basis of $V$ then pick $w_j = Tv_j$, the $w_j$'s will be a basis for $W$ since $T$ is invertible meaning $\mathcal M(T)$ with respect to these bases is the identity.