The easier part is to see that $\lim_{\infty \leftarrow n} H^*(G(d,n)) \cong \mathrm{Rep}(GL_d)$ as rings. The representation ring of $GL_d$ is the same as the $GL_d$-equivariant $K$-theory of a point. Let $\mathrm{Mat}_{d \times n}^{\circ}$ be the full rank $d \times n$-matrices. I won't define this rigorously, but the ind-scheme $\lim_{n \to \infty} \mathrm{Mat}_{d \times n}^{\circ}$ is "contractible", so the $GL_d$-equivariant $K$-theory of a point is $K^0(\lim_{n \to \infty} \mathrm{Mat}_{d \times n}^{\circ} / GL_d)$. One can justify turning this into $\lim_{n \leftarrow \infty} K^0( \mathrm{Mat}_{d \times n}^{\circ} / GL_d)$. Of course, $\mathrm{Mat}_{d \times n}^{\circ} / GL_d = G(d,n)$.` The Chern character map is an isomorphism between $K^0(G(d,n))$ and $H^*(G(d,n))$, at least once we tensor with $\mathbb{Q}$.

The hard thing is to explain why there should be an isomorphism which takes irreps to Schubert cycles, and works over $\mathbb{Z}$. This is especially difficult because Chern character is not that isomorphism. I don't know any really short answer to this question, but Harry Tamvakis has an expository paper which does a pretty good job.

The easier part is to see that $\lim_{\infty \leftarrow n} H^*(G(d,n))$ is very close to $\mathrm{Rep}(GL_d)$ as rings. (As Allen points out below, we don't want all the representations of $GL_d$, only the polynomial ones. I'll continue glossing over that point, but it is another sign that any answer has to be somewhat complex.)

The representation ring of $GL_d$ is the same as the $GL_d$-equivariant $K$-theory of a point. Let $\mathrm{Mat}_{d \times n}^{\circ}$ be the full rank $d \times n$-matrices. I won't define this rigorously, but the ind-scheme $\lim_{n \to \infty} \mathrm{Mat}_{d \times n}^{\circ}$ is "contractible", so the $GL_d$-equivariant $K$-theory of a point is $K^0(\lim_{n \to \infty} \mathrm{Mat}_{d \times n}^{\circ} / GL_d)$. One can justify turning this into $\lim_{n \leftarrow \infty} K^0( \mathrm{Mat}_{d \times n}^{\circ} / GL_d)$. Of course, $\mathrm{Mat}_{d \times n}^{\circ} / GL_d = G(d,n)$.` 

The Chern character map is an isomorphism between sends $K^0(G(d,n))$ to the completion of $H^*(G(d,n))$, at least once we tensor with $\mathbb{Q}$. (That completion is related to the issue of whether we work with polynomial representations or all representations.)

The hard thing is to explain why there should be an isomorphism which takes irreps to Schubert cycles, and works over $\mathbb{Z}$. This is especially difficult because Chern character is not that isomorphism. I don't know any really short answer to this question, but Harry Tamvakis has an expository paper which does a pretty good job.