Here it will be shown how conditions on the f_n's or on f can be modified to make the sequence (x_n) convergent.
First of all, the condition that the f_n's be twice differentiable is of no help; so, this condition will be dropped.
Suppose that for each n there exists some x_n\in\mathbb{R} such that:
f_n is semi-strongly convex on (-\infty,x_n) in the sense that f_n is convex on (-\infty,x_n) and f_n(x)/|x|\to\infty as x\to-\infty;
f_n is semi-strongly concave on (x_n,\infty) in the sense that f_n is concave on (x_n,\infty) and f_n(x)/x\to-\infty as x\to\infty.
Note that, if f_n is strongly convex on (-\infty,x_n), then f_n is semi-strongly convex on (-\infty,x_n); similarly, if -f_n is strongly convex on (x_n,\infty), then f_n is semi-strongly concave on (x_n,\infty).
Suppose also that (f_n) uniformly converges to a function f, which
is not affine on any nonempty interval.
Then the sequence (x_n) is convergent.
Indeed, suppose first that the sequence (x_n) is unbounded. Then, using a left-right symmetry and passing to a subsequence, without loss of generality (wlog) assume that x_n\to\infty. It follows that the limit function f is convex on \mathbb R. So, f(x)\ge a+bx for some real a,b and all real x. Therefore and in view of semi-strongly concavity of f_n on (x_n,\infty), for each n we have f_n(x)-f(x)\to-\infty as x\to\infty, which contradicts the uniform convergence of (f_n) to f.
So, the sequence (x_n) is bounded. Take any subsequence (x_{n_k}) of the sequence (x_n) converging to a limit x_*. Then f is convex on (-\infty,x_*) and concave on (x_*,\infty). Since f is not affine on any nonempty interval, the point x_* is uniquely determined by f. So, any converging subsequence of the sequence (x_n) converges to the same limit x_*. Since the sequence (x_n) is bounded, we conclude that x_n\to x_*, as claimed.
This reasoning shows that the following is true as well:
Suppose that for each n there exists some x_n\in\mathbb{R} such that:
f_n is convex on (-\infty,x_n);
f_n is concave on (x_n,\infty).
Suppose also that (f_n) converges pointwise (not necessarily uniformly) to a function f, which
is not affine on any nonempty interval, not convex on \mathbb R, and not concave on \mathbb R.
Then the sequence (x_n) is convergent.
Details on the latter assertion, requested by Gaetano: Suppose first that the sequence (x_n) is unbounded. Then, using a left-right symmetry and passing to a subsequence, wlog assume that x_n\to\infty. It follows that the limit function f is convex on \mathbb R, which contradicts the assumption that f is not convex on \mathbb R.
So, the sequence (x_n) is bounded. Take any subsequence (x_{n_k}) of the sequence (x_n) converging to a limit x_*. Then f is convex on (-\infty,x_*) and concave on (x_*,\infty). Since f is not affine on any nonempty interval, the point x_* is uniquely determined by f. So, any converging subsequence of the sequence (x_n) converges to the same limit x_*. Since the sequence (x_n) is bounded, we conclude that x_n\to x_*, as claimed.