Pullback of sheaves. Let $ f : X \to Y$ be a continuous map of topological spaces.. Let $ \mathcal{F} = \underline{\mathbb{R}} $ be a sheaf of locally constant function from Y to $\mathbb{R}$ on Y. Let $ f^* \mathcal{F}$ be the pullback of the sheaf on X. Is the pullback of $ \mathcal{F}$, the sheaf of locally constant functions on X?
The answer is yes. This is one of the cases where it is better to think as sheaves as étalé spaces.   

The constant sheaf $\mathcal F=\mathbb R_Y$on $Y$ with stalk $\mathbb R$ corresponds to the étalé space $Y\times \mathbb R_{disc}$, where $\mathbb R_{disc}$ denotes $\mathbb R$ endowed with the discrete topology.
The étalé space corresponding to $f^*(\mathcal F)$ is the fibre product $X\times_Y (Y\times \mathbb R_{disc})$ and this fibre product is homeomorphic to $X\times \mathbb R_{disc}$.
Now, the étalé space $X\times \mathbb R_{disc}$ on $X$ corresponds to the constant sheaf $\mathbb R_X$ and this  shows what you wanted to know: $$  f^*(\mathcal F) =f^*(\mathbb R_Y) =     \mathbb R_X      $$
No, suppose that $Y$ is a point. The pullback $f^*{\cal F}$ is a constant sheaf, and the sheaf of locally functions defined on $X$ is not always the constant sheaf.