What you have written above isn't classical Satake; it's the generalized Bruhat decomposition. Classical Satake is a much more interesting theorem, which says that the Hecke algebra of $G(\mathcal{K})$ over $G(\mathcal{O})$ (the compactly supported $G(\mathcal{O})$ bi-invariant functions on $G(\mathcal{K})$ with convolution multiplication) is isomorphic to the representation ring of $G^\vee$.
Why is this interesting? Because the Hecke algebra $L^2[G(\mathcal{O})\backslash G(\mathcal{K})/G(\mathcal{O})]$ is the endomorphism algebra of $L^2(G(\mathcal{K})/G(\mathcal{O}))$, so there's a bijection between $W$-orbits on $T^\vee$ and representations of $G(\mathcal{K})$ appearing in the $L^2$ above.
(1) Short answer to first question: $T_p$ is about $p$-isogenies, and in char. $p$ there is a canonical $p$-isogeny, namely Frobenius.
Details:
The Hecke correspondence $T_p$ has the following definition, in modular terms:
Let $(E,C)$ be a point of $X_0(N)$, i.e. a modular curve together with a cyclic subgroup
of order $N$. Now $T_p$ (for $p$ not dividing $N$) is a correspondence (multi-valued function) which maps
$(E,C)$ to $\sum_D (E/D, (C+D)/D)$, where $D$ runs over all subgroups of $E$ of degree $p$.
(There are $p+1$ of these.)
Here is another way to write this, which will work better in char. $p$:
map $(E,C)$ to $\sum_{\phi:E \rightarrow E'}(E',\phi(C)),$
where the sum is over all degree $p$ isogenies $\phi:E\rightarrow E'.$ Giving a degree
$p$ isogeny in char. 0 is the same as choosing a subgroup of order $D$ of $E$ (its kernel),
but in char. $p$ the kernel of an isogeny can be a subgroup scheme which is non-reduced,
and so has no points, and hence can't be described simply in terms of subgroups of points.
Thus this latter description is the better one to use to compute the reduction of the
correspondence $T_p$ mod $p$.
Now if $E$ is an elliptic curve in char. $p$, any $p$-isogeny $E \to E'$ is either
Frobenius $Fr$, or the dual isogeny to Frobenius (often called Vershiebung).
Now Frobenius takes an elliptic curve $E$ with $j$-invariant $j$ to the elliptic curve
$E^{(p)}$ with $j$-invariant
$j^p$. So the correspondence on $X_0(N)$ in char. $p$ which maps $(E,C)$ to $(E^{(p)},
Fr(C))$ is itself the Frobenius correspondence on $X_0(N)$. And the correspondence
which maps $(E,C)$ to its image under the dual to Frobenius is the transpose
to Frobenius (domain and codomain are switched). Since there are no other $p$-isogenies in char. $p$
we see that $T_p$ mod $p = Fr + Fr'$ as correspondences on $X_0(N)$ in char. $p$;
this is the Eichler--Shimura relation.
(2) Note that only weight 2 eigenforms with rational Hecke eigenvalues give elliptic curves;
more general eigenforms give abelian varieties.
An easy computation shows that if $f$ is a Hecke eigenform, than the $L$-funcion
$L(f,s)$, obtained by Mellin transform, has a degree 2 Euler product. A more conceptual answer would probably involve describing how automorphic representations
factor as a tensor product of local factors, but that it a very different topic from Eichler--Shimura, and I won't say more here.
Best Answer
Let $M$ be a closed orientable $n$-manifold containing the compact set $X$. Given an $n-q-1$-cocyle on $X$ (I am choosing this degree just to match with the notation of the Wikipedia article to which you linked), we extend it to some small open neighbourhood $U$ of $X$. By Lefschetz--Poincare duality on the open manifold $U$, we can convert this $n-q-1$-cocylce into a Borel--Moore cycle (i.e. a locally-finite cycle made up of infinitely many simplices) on $U$ of degree $q+1$. Throwing away those simplices lying in $U \setminus X$, we obtain a usual (i.e. finitely supported) cycle giving a class in $H_{q+1}(U,U\setminus X) = H_{q+1}(M,M\setminus X)$ (the isomorphism holding via excision). Alexander duality for an arbitrary manifold then states that the map $H^{n-q-1}(X) \to H_{q+1}(M,M \setminus X)$ is an isomorphism. (If $X$ is very pathological, then we should be careful in how define the left-hand side, to be sure that every cochain actually extends to some neighbourhood of $X$.)
Now if $M = S^{n+1}$, then $H^i(S^{n+1})$ is almost always zero, and so we may use the boundary map for the long exact sequence of a pair to identify $H_{q+1}(S^{n+1}, S^{n+1}\setminus X)$ with $H_{q}(S^{n+1}\setminus X)$ modulo worrying about reduced vs. usual homology/cohomology (to deal with the fact that $H^i(S^{n+1})$ is non-zero at the extremal points $i = 0$ or $n$).
So, in short: we take a cocycle on $X$, expand it slightly to a cocyle on $U$, represent this by a Borel--Moore cycle of the appropriate degree, throw away those simplices lying entirely outside $X$, so that it is now a chain with boundary lying outside $X$, and finally take this boundary, which is now a cycle in $S^{n+1} \setminus X$.
(I found these notes of Jesper Moller helpful in understanding the general structure of Alexander duality.)
One last thing: it might help to think this through in the case of a circle embedded in $S^2$. We should thicken the circle up slightly to an embedded strip. If we then take our cohomology class to be the generator of $H^1(S^1)$, the corresponding Borel--Moore cycle is just a longitudinal ray of the strip (i.e. if the strip is $S^1 \times I$, where $I$ is an open interval, then the Borel--More cycle is just $\{\text{point}\} \times I$).
If we cut $I$ down to a closed subinterval $I'$ and then take its boundary, we get a pair of points, which you can see intuitively will lie one in each of the components of the complement of the $S^1$ in $S^2$.
More rigorously, Alexander duality will show that these two points generate the reduced $H^0$ of the complement of the $S^1$, and this is how Alexander duality proves the Jordan curve theorem. Hopefully the above sketch supplies some geometric intuition to this argument.