Basically you want to construct a chain of inequalities that links the smaller expression to the larger expression. Induction is often helpful in these cases.
A useful theorem for Knuth arrows is $(a \uparrow^n b) \uparrow^n c < a \uparrow^n (b+c)$, proven in this paper. It is also proven that $a \uparrow^n c$ is monotonic in $a,n$, and $c$ when $a,c \ge 3$, which is useful as well.
For example, one can easily see that $S < 3 \uparrow\uparrow 6$, so
$$S^{S^{S^\cdots}} = S \uparrow \uparrow S < (3\uparrow\uparrow 6)\uparrow\uparrow(3 \uparrow\uparrow 6) < 3\uparrow\uparrow (6 + 3\uparrow\uparrow 6) < 3 \uparrow\uparrow (3 \uparrow\uparrow 7) < 3 \uparrow\uparrow (3\uparrow\uparrow 3^{3^3}) = 3 \uparrow\uparrow (3\uparrow\uparrow\uparrow 3) = 3\uparrow\uparrow\uparrow 4 < 3\uparrow\uparrow\uparrow (3 \uparrow\uparrow\uparrow 3) = 3\uparrow\uparrow\uparrow\uparrow 3 = G_1$$
To address your harder question, first we need to know what $G(0,y)$ is. Since we need $G(0,3) =4$ so that $G(64,3)$ is Graham's number, I will assume that $G(0,y)=4$.
Theorem: $G(n,S) < G(n+1,3)$
We will prove this by induction. First, observe that $G(0,S) = 4 < 3\uparrow\uparrow\uparrow\uparrow 3 = G(1,3)$.
Observe that for $n \ge 3$,
$$S \uparrow^n S < (3\uparrow\uparrow 6)\uparrow^n (3\uparrow\uparrow 6) < (3\uparrow^n 6)\uparrow^n (3\uparrow\uparrow 6) < 3\uparrow^n (6+3\uparrow\uparrow 6) < 3\uparrow^n (3\uparrow\uparrow\uparrow 3) \le 3\uparrow^n (3\uparrow^n 3) = 3\uparrow^{n+1} 3$$
So if we have $G(n,S) < G(n+1,3)$, then $G(n,S)+1 \le G(n+1,3)$, so
$$G(n+1,S) = S \uparrow^{G(n,S)} S < 3 \uparrow^{G(n,S)+1} 3 \le 3 \uparrow^{G(n+1,3)} 3 = G(n+2,3)$$
and the theorem follows by induction.
So in particular, $G(60,S) < G(61,3) < G(64,3)$.