It's probably easiest to see this by example. Let's take a quadric curve in C^2 defined by
x^2 + y^2 - 1 = 0
which we cill call V.
Polynomials on C^2 are given by the ring C[x,y]. What about polynomials on V? Clearly, any polynomial on C^2 can be restricted to a polynomial on V, but this restriction isn't injective. For example, the polynomial x^2 + y^2 and the polynomial 1 both take the same value on V (hence are the same polynomial on V) but are clearly different when considering all of C^2 .
More generally, two polynomials p(x,y) and q(x,y) are the same on V if their difference is zero on V. This means (by some basic AG theorem) that
p(x,y)-q(x,y) = f(x,y)(x^2 + y^2 - 1)
i.e. p(x,y)-q(x,y) lies in the ideal I(V).
Conversely, take a polynomial p(x,y) defined only on V. We want to write down a formula for this polynomial in terms of x and y, so we want to express it as a polynomial on all of C^2. As we have seen, we have many inequivalent choices of expressions for p(x,y), but all of them reduce to the same polynomial when restricted to V.
Hence, if we're talking about p(x,y) as a function on V, it is precise to instead refer to the set of all polynomials in two variables that restrict to the same function on V. If we want a formula, we may pick one from this set (the "representative"), but all choices are equivalent on V.
(This is why some people write p(x,y) for the polynomial on C^2 , and p(x,y) + I(V) for the polynomial on V, since any expression of the form p(x,y)+ i(x,y) for i in I(V) restricts to the same polynomial on V)