What is the math behind the game Spot It?

The celebrated Ray-Chaudhuri–Wilson theorem states that $C \leq S$, contradicting your numbers.

An almost matching construction is as follows. Pick some prime number $n$. Our universe, of size $n^2+n+1$, consists of pairs of numbers in $\{0,\ldots,n-1\}$ plus $n+1$ singletons $\{0,1,\ldots,n-1,\infty\}$ ("points at infinity"). For each $0 \leq a \leq n-1$ and $0 \leq b \leq n-1$ we will have a card of size $n+1$ containing the pairs $\{(x,ax+b \mod{n})\}$ and the singleton $a$. There are also $n$ special cards, for each $0 \leq c \leq n-1$, containing the pairs $\{(c,x)\}$ and the singleton $\infty$. One super special card contains all $n+1$ singletons.

Clearly two cards with the same $a$ intersect only at the singleton. Two cards with different $a$s intersect at the unique solution to $a_1x+b_1 = a_2x+b_2 \pmod{n}$. Two special cards intersect only at the singleton, and a normal and a special card intersect at $(c,ac+b)$. Finally, the super special card intersects the rest at a singleton.

In total, we have $n^2+n+1$ cards and symbols, each card containing $n+1$ symbols, and two cards intersecting at exactly one symbol. In your case $n=7$ and so the number of cards and symbols should be $7^2+7+1 = 57$.

Here's an article (in French) that aims to explain the mathematics behind the game to a wide audience.

In the interest of link rot prevention, here are two diagrams from the article that may be of interest even to non-French speakers:

I came to the conclusion it must be $57$ or more symbols the following simple way: the total number of symbols shown on all cards is $55\times8=440$. If it was $50$ different symbols only, each must be shown $440:50=8.8$ times, i.e. some at least $9$ times.

If you took the $9$ cards with one common symbol, all the other symbols would need to be different, i.e. you'd need $(8-1)\times9+1=64$ different symbols.

If we reduce the max. usage per symbol to $8$, you only need $(8-1)\times8+1=57$ Symbols. As $57\times8=456$, this also exceeds the #of symbols shown thus being a valid solution.

To be able to use fewer symbols (e.g. $56$), usage per symbol would require reduction to 7 per (each individual) symbol, which would limit the no. of cards to $56\times7:8=49$.

Thus, with $55$ cards, the minimum number of different symbols is $57$.