A: set of all row vectors: ($a_1$, ...,$a_n$), $a_j$ in K; addition, multiplication defined componentwise.
B: set of all functions with values in K, defined on an arbitrary set S.
C: set of all polynomials of degree less than n with coefficients in K.
To show A and C are isomorphic, can I construct a function $f(x)=a_1+a_2 x+a_3 x^3$+...+$a_n x^{n-1}$, and show f is a one-to-one correspondence?
To show that "if S has n elements, then A and B are isomorphic," can I define S as (1, x, $x^2$, ...,$x^{n-1}$) and apply the result that A and C are isomorphic?
You have the right idea, though I'd write it a little bit more carefully. We can't "define $S$" as we wish because $S$ is given to us. We can, however, argue as follows. Let $S = \{s_0, s_1, \dots, s_{n-1}\}$. Note that any function $S \to K$ is characterized by $n$ values $f(s_0), f(s_1), \dots, f(s_{n-1})$. Now we build an isomorphism between $B$ and $C$ by assigning to each function in $B$ the polynomial $f(s_0) + f(s_1)x + f(s_2)x^2 + \dots + f(s_{n-1})x^{n-1}$. The assignment is a bijection because given any polynomial in $C$, you can read the coefficients and use them as outputs of a function on $S$. You have to confirm that scalar multiples and sums of functions correspond to scalar multiples and sums of polynomials. Then once you've confirmed the bijection is indeed an isomorphism of vector spaces, you can argue as you did that since $A \cong C$ and $B \cong C$, $A \cong B$.