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I am not sure whether this counts as heuristics, as it goes even deeper into free probability results, but it might give some high-level kind of idea why this result should be true. The main fact about a unitarily-invariant matrix $X=U^*AU$ is that asymptotically the free cumulants $\kappa_n(X)$ of its eigenvalue distribution are given by scaled limits of classical cumulants $c_n$ of the entries $x_{ij}$ of the matrix, with a cycle structure in the indices, namely $$\kappa_n(X)=\lim_{N\to\infty} N^{n-1} c_n(x_{i_1i_2},x_{i_2i_3},\dots,x_{i_ni_1}).$$ The actual value of the $i_k$ does not matter, because asymptotically they all have the same leading order, so we might just take $c_n(x_{12},x_{23},\dots,x_{n1})$. This fact is surely not obvious and buried in the literature; it can be found, e.g., in Theorem 2.6 of this paper. Taking this for granted then the statement about the free compression is obvious. Take as new matrix $Y$ the upper $\tau N\times \tau N$-corner of our $N\times N$-matrix $X$. Then $Y$ is still unitarily-invariant, so we get its free cumulants as the limits of classical cumulants in its entries, but for the entries we can take the same cycle as for $X$. So we have $$\kappa_n(Y)=\lim_{N\to\infty} (\tau N)^{n-1} c_n(x_{i_1i_2},x_{i_2i_3},\dots,x_{i_ni_1})= \frac 1\tau \lim_{N\to\infty} N^{n-1} c_n(\tau x_{i_1i_2}, \tau x_{i_2i_3},\dots,\tau x_{i_ni_1})= \frac 1\tau \kappa_n(\tau X).$$ By realizing the free convolution as the sum of independent unitarily-invariant random matrices, one sees that one has the same result also in this case (by taking into account that mixed classical cumulants in independent variables vanish).