Inspired by @obsolesced, it is simple to find solutions based on any set that is mapped to itself by f1 and f2. For example, let

S = {q1 + q2 Sqrt[2]}  for {q1, q2} rational

then

f[x] = f1[x]   x in S
f[x] = f2[x]   x not in S

satisfies the required functional relation. It is continuous nowhere (except x = -7/2) and both S and its complement are uncountably infinite.

It would be interesting to know if there are any solutions to the functional equation with f[x] not in {f1[x],f2[x]}.

Inspired by @obsolesced, it is simple to find solutions based on any set that is mapped to itself by f1 and f2. For example, let

S = {q1 + q2 Sqrt[2]}  for {q1, q2} rational

then

f[x] = f1[x]   x in S
f[x] = f2[x]   x not in S

satisfies the required functional relation. It is continuous nowhere (except x = -7/2) and S is countably infinite and its complement uncountably infinite.

Alternatively, make S the Algebraic numbers, then its complement is the Transcendentals - both uncountably infinite.

It would be interesting to know if there are any solutions to the functional equation with f[x] not in {f1[x],f2[x]}.