Let the two points of symmetry be $(p,f(p)), (q,f(q))$. $f(p+(x-q)) + f(p-(x-q)) = 2f(p)$ and $f(q+(x-p)) + f(q-(x-p)) = 2f(q)$. Since $f(p-(x-q)) = f(p+q-x) = f(q-(x-p))$, we have $f(q+(x-p)) = 2f(q) - 2f(p) + f(p+(x-q))$, so $f(x + (q-p)) = 2f(q) - 2f(p) + f(x + (p-q))$. Substituting $x = y+q-p$ gives $f(y + (2p-2q)) = 2f(q) - 2f(p) + f(y)$. Let $g(x) = \dfrac{f(q)-f(p)}{p-q} x$, $h$ be a periodic function of period $2p-2q$, $f(x) = g(x) + h(x)$. Observe that the desired preposition is still true for all $x$, so $g,h$ are the two functions we're looking for.

We are simply looking for an $a,b$, such that, $f(x)-(ax+b)$ is periodic; or equivalently, searching $T,a,b$ such that, $f(x+T)-(a(x+T)+b)=f(x)-(ax+b)\iff f(x+T)-f(x)=\text{constant}$ for every $x$. Now, note that, $$ f(a-x)+f(a+x)=2f(a)\implies f(b-x)+f(2a-b+x)=2f(a)\implies 2f(b)-f(b+x)+f(2a-b+x)=2f(a). $$Upon substituting $x\mapsto x-b$, we get, $f(2a-2b+x)-f(x)=2(f(a)-f(b))$, showing the desired claim.