Suppose that ZFC is consistent, and let ZFC'=ZFC+Con(ZFC). Can one construct two statements $\phi_1$ and $\phi_2$ such that
$$ ZFC' \vdash ((ZFC \vdash \phi_1) \ \text{or} \ (ZFC \vdash \phi_2)) $$
$$ ZFC' \not\vdash (ZFC \vdash \phi_1), \ \text{and} \ ZFC' \not\vdash (ZFC \vdash \phi_2) $$
Hint: $ZFC'$ still has an $\omega$-model.
Here is a more complete answer.
The statement $\Gamma \equiv (ZFC \vdash \phi_1) \lor (ZFC \vdash \phi_2)$ is provably equivalent in ZFC to the $\Sigma^0_1$ statement "there is an $n \in \omega$ which is either a coded proof of $\phi_1$ or a coded proof of $\phi_2$". Because $ZFC'$ has an $\omega$-model, if ZFC' proves $\Gamma$ then there really is such an $n$, in which case $ZFC'$ proves either "n is a coded proof of $\phi_1$" or "$n$ is a coded proof of $\phi_2$" (because both of those are bounded-quantifier sentences, they are provable if they are true). So the answer is that one cannot construct statements as described in the question.
For generality, we could replace $ZFC'$ in $\Gamma$ with any sufficiently strong theory that has an $\omega$-model, and replace $ZFC$ in $\Gamma$ with any effective theory.