Suppose that $X$ is a (real) Hausdorff topological vector space and let $k \in (0,\infty)$.
Let $f,g\in C(X)$ and suppose that $Im(f)=Im(g)=[0,\infty)$. So in particular, both have minimum at $0$ and I assume that that $g(0)=0=f(0)$. Fix some $y \in X$. Does there exist some $x\in X$ such that $ f(x)+g(y-x)=k ? $
If not, are there mild conditions under which we can guarantee it?
Well, consider the function $h:X\to\mathbb{R}$ defined by $h(x)=f(x)+g(y-x)$. Then $h$ is continuous, so its image is an interval since $X$ is connected. Also, $h(0)=g(y)$, $h(y)=f(y)$, and $h$ is unbounded each term of $h(x)$ can get arbitrarily large and the other term is nonnegative. So, the image of $h$ contains $[a,\infty)$ where $a=\min(f(y),g(y))$ (so the answer to your question is yes if $k\geq a$).
In general, though, the image of $h$ may not be any larger than $[a,\infty)$. For instance, suppose $f=g$ is a norm on $X$. Then $h(x)=f(x)+f(y-x)\geq f(y)$ for all $x$ by the triangle inequality.