Let $X$ be a topological space. By a morphism of base-$X$ sheaves, we mean a morphism of sheaves on $X$, whose underlying base space map is the identity on $X$.
We will use a stronger notion of a vector bundle, which we refer to as a "TVS bundle": this is a vector bundle with a topological vector space (TVS) structure on each fiber, and the "typical fiber" (used in the trivializations) is a TVS, and the local trivializations are, when restricted to each fiber, continuous linear with continuous inverse.
Suppose $F,F'$ are sheaves of real vector spaces on $X$, and suppose each of $F,F'$ is the sheaf of sections of some TVS bundle $E,E'$ (not necessarily of finite rank) on $X$.
My question is, does every morphism $F \rightarrow F'$ of base-$X$ sheaves (of real vector spaces) arise from a morphism $E \rightarrow E'$ of TVS bundles over $X$ (lying over the identity on $X$)? If yes: what tools are used to show this; and, is the morphism $E \rightarrow E'$ uniquely determined?
Do you mean for $F$ and $F'$ to be sheaves of topological real vector spaces on $X$? Otherwise the answer is no for trivial reasons:
Take $X$ to be the one-point space. Let $E = E'$ be an infinite-dimensional normed vector space, viewed as a TVS bundle on $X$. Let $f : E \to E'$ be any linear function which is not continuous. Then $F = F'$ is the underlying vector space of $E = E'$ (viewed as a sheaf of vector spaces on $X$) and $f : F \to F'$ is a well-defined morphism of sheaves, but $f$ does not arise from a morphism of TVS bundles $E \to E'$ (since it is not continuous).