Proof:

Invoking the uniform continuity of $f$ with $\varepsilon=2$, there exists $\delta > 0$ such that

$$\Vert x-y\Vert < \delta \Rightarrow \vert{ f(x) - f(y)}\vert < 2\,$$

which in turn implies that $f(x) \neq -f(y).$ Now choose $n\in \mathbb{N}$ such that $n^{-1}\Vert x\Vert < \delta$ for all $x \in X$ which is possible since $X$ is compact. This $n$ is now fixed for the rest of the discussion. For $x \in X$ and $j=0,1,\dots ,{n-1}$

$$\Big\Vert\frac{j}{n}x - \frac{(j+1)}{n}x\Big\Vert < \delta,$$

whereby,

$$f(\frac{j+1}{n}x) \neq -f(\frac{j}{n}x).$$

From this we conclude that the function given by

   lg$$\Big(\frac{f(\frac{j+1}{n}x)}{f(\frac{j}{n}x)}\Big),\quad x \in X$$

is continuous with respect to $x$. We now claim that

$${\tilde f}(x) = t_0 + \sum_{j=0}^{n-1}$$   lg$$\Big(\frac{f(\frac{j+1}{n}x)}{f(\frac{j}{n}x)}\Big)$$

is the required continuous function. Observe that lg$(1) =0,\; \tilde{f}(0)=t_0$ and

   ex$${\tilde f}(x) = ($$ex$$(t_0)) \cdot \frac{f(\frac{1}{n}x)}{f(0)} \cdot \frac{f(\frac{2}{n}x)}{f(\frac{1}{n}x)} \cdots \frac{f(\frac{n}{n}x)}{f(\frac{n-1}{n}x)} = f(x).$$

Turning to the proof of uniqueness of the lift ${\tilde f}$, suppose ${\tilde f}_1, {\tilde f}_2 : X \longrightarrow \mathbb{R}$ are two continuous functions such that ${\tilde f}_1(0) = {\tilde f}_2(0) = t_0$ and ex${\tilde f}_1(x) =$   ex${\tilde f}_2(x) = f(x).$ Then ex$({\tilde f}_1(x) - {\tilde f}_2(x)) = 1,$ which implies ${\tilde f}_1(x) - {\tilde f}_2(x) \in \mathbb{Z}$ (see note below). Since both functions are continuous, agree at the origin and $X$ is connected, we conclude that

$${\tilde f}_1(x) \equiv {\tilde f}_2(x). \eqno\square$$