The accepted answer to this question is very close to yet still missed a bit to what I want to know. It gives an example where the minimizer of $f=f_1+f_2$ is not a convex combination of the minimizers of $f_1$ and $f_2$. However, what happens of $f$ itself is a convex combination of $f_1$ and $f_2$ (i.e., $f=\theta f_1 + (1-\theta) f_2$ where $0<\theta<1$)? Is the statement 'the minimizer of $f$ is a convex combination of the minimizers of $f_1$ and $f_2$' true then? Furthermore, could the statement 'the minimizer of $f$ is the weighted average of the minimizers of $f_1$ and $f_2$ weighted according to $\theta$ and $1-\theta$' be true (needless to say this is even harder to be true)? Finally, if neither of those two statement is true, my guess is the following statement has to be true: given d-dimensional convex functions $f_1,...,f_n$, and the coordinates of their respective minimizer $(m_{1,1},...,m_{1,d}),...,(m_{n,1},...,m_{n,d})$, the coordinates of the minimizer of the convex combination of $f_1,...,f_n$ is $(m_{1}^*,...,m_{d}^*)$ such that $ min_{i}\{m_{i,j}\} \leq m_{j}^*\leq max_{i}\{m_{i,j}\} \,\forall j \in [d]$. But I'm not sure how to prove it.
The accepted answer here is related to my last statement, but it considered sum of $f_1$ and $f_2$ instead of their convex combination.
[Comment elevated to partial answer at suggestion of OP]
To answer part of the question: OP refers to an example where the minimizer of $f=f_1+f_2$ is not a convex combination of the minimizers of $f_1$ and $f_2$. If we let $$ g={f_1+f_2\over2} $$ then $g$ is a convex combination of $f_1$ and $f_2$, and the minimizer of $g$ is the same as the minimizer of $f$ and thus not a convex combination of the minimizers of $f_1$ and $f_2$.