I don't have
Sorry, the relevant sources in front first edition of me, so take this with a grain of saltanswer was shamefully incoherent. We'll see if this attempt is any better.
Any double coset KgK (for G and K as given) has a unique representative in elementary divisor form $\binom{a0}{0d}$ where a and d are (possibly negative) powers of p and a/d is an a p-adic integer (i.e., a positive power of p). The Hecke operator $T(p^n)$ is given by a sum over convolutions with KgK as g ranges over elementary divisor matrices with p-adic integer entries with determinant $p^n$. In particular, T(p) is given by convolving with the double coset corresponding to $\binom{p0}{01}$. In the notation of Buzzard's answer, the operators $T(p^n)$ generate a the subalgebra of the Hecke algebra . It is generated by $S = \binom{p0}{0p}$ and $T = \binom{p0}{01}$, and it coincides with the subalgebra that is generated by convolution with those double cosets whose elementary divisor representative has p-adic integer entries.
You can find a non-adelic treatment in terms of Hecke operators acting on modular forms on the upper half plane in section 1.4 of Bump's Automorphic Forms and Representations, where he introduces operators $T_\alpha$ for diagonal matrices $\alpha$ in elementary divisor form, and shows how $T(n)$ is given as a sum over double cosets with determinant n. Decomposing these double cosets into left cosets for $\Gamma(1)$ yields the usual set of representatives $\{ a,b,d| ad=n, 0 \leq b < d \}$ over which one sums when evaluating a Hecke operator.
The Satake isomorphism gives an isomorphism with the representation ring of $GL_2(\mathbb{C})$, which is commutative and finitely generated. Searching for "Satake isomorphism" yielded This implies the Hecke algebra here is commutative and finitely generated, but this rather nice introduction by Grosscan be seen without invoking such machinery. In the case of $GL_2$, the isomorphism can be made very explicit. $T(p)$ corresponds to the standard 2 dimensional representation, the scalar matrices give powers of determinant, and $T(p^n)$ corresponds to the $n$th symmetric power.
