if you define the following map:
p(x)--> (p(x)+I, p(x)+J) where I=(f(x)), J=(g(x)) and p(x) \in k[x]
you can prove that it is a ring epimorphism with kernel = I-intersection-J which is equal to IJ =(f(x)g(x)) since I+J = (f(x), g(x)) = k[x] and then you can apply the 1st isomorphism theorem for rings.
exactly!
I think that this is the most difficult part of the proof.
you can work as follows:
there exist s(x) and t(x) such that 1 = s(x)f(x)+t(x)g(x), since 1 \in k --> 1\in k[x] --> 1\in (f(x), g(x)).
if p(x) = a(x)t(x)g(x)+b(x)s(x)f(x) then you can verify that p(x)+I = a(x) +I and p(x)+J = b(x) +J
so (p(x)+I, p(x)+J) =(a(x)+I, b(x)+J)
with (a(x)+I, b(x)+J) an arbitrary element \in k[x]/(f(x)) \bigoplus k[x]/(g(x)) .