I may have found a quicker solution.
Let L be the limit of f + g. Since we can add -L to g or to f without changing T1 or T2, we can assume that L = 0.
Since T1 and T2 are not 0, for all real number x, f(x + nT1) + g(x + nT1) --> 0 when n goes to infinty, so we obtain that g(x+nT1) --> -f(x) when n goes to infinity. With the same idea, we get that f(x+nT2) --> -g(x) when n goes to infinity. Combing those two small resultats, we get that
f(x + n(T1 + T2)) + g(x + n(T1+T2)) --> 0 when n goes to infinity, but also
f(x + n(T1 + T2)) + g(x + n(T1+T2)) --> -f(x) - g(x) when n goes to infinity.
So we obtain that f(x) = -g(x) for all x, so f+g is 0, and we are done. Quicker, but slightly not pedagogical.