Find MST on grid graph with only weight of 1 and 2 in $O(|V|+|E|)$
Let $n=|V|$ and $m=|E|$.
Intuitively you want the to return the union of the edges in 1) a maximal spanning forest $F$ of the graph induced by the edges of weight $1$, with 2) a maximal spanning forest $F'$ of the graph obtained by identifying the edges of each tree in $F$ into a single vertex (where each edge in $F'$ actually represents an edge of $G$).
Some care is required to attain a running time of $O(n)$. The details are as follows.
 
Let $G_1$ be subgraph of $G$ induced by the edges of weight $1$.
Let $C_1, \dots, C_k$ be the connected components of $G_1$.
For each $C_i$, compute any a spanning tree $T_i = (V_i, E_i)$ of $C_i$.
This requires $O(m)=O(n)$ time in total.
For each edge $e=(u,v)$ of weight $2$ in $G$ let $i$ and $j$ be such that $u \in C_i$ and $v \in C_j$. Let $k(e) = (\min\{i,j\}, \max\{i,j\})$.
Sort the edges of $G$ in nondecresing order of $k(\cdot)$, keep at most one edge for each value of $k(\cdot)$. Let $S$ the resulting ordered set of edges.
Notice that $S$ can be found in time $O(m)=O(n)$ using radix-sort.
Create a graph $G_2$ with vertex set $\{1, \dots, k\}$ and edge set $\{ k(e) : e \in S \}$. For an edge $(i,j)$ in $G_2$ let $\ell(i,j)$ be and edge $e$ in $G$ such that $k(e) = (i,j)$.
This label $\ell(i,j)$ can be stored along with the edge $(i,j)$ itself, so that given $(i,j)$ we can find $\ell(i,j)$ in $O(1)$ time.
This step also requires $O(n)$ time.
Finally, compute any spanning tree $T' = (V', E')$ of $G_2$.
An MST of $G$ is the tree induced by the edges in $\left( \bigcup_{i=1}^k E_i \right) \cup \{ \ell(i,j) \mid (i,j) \in E'\}$.
Overall, the whole algorithm takes $O(n)$ time.
The same algorithm extends naturally to any constant number of distinct edge weights.
Here is a visualization of the algorithm on the graph you proposed in the comments. Blue edges has weight $1$, red edges have weight $2$. The connected components of $G_1$ are highlighted in gray (and each connected component will be represented by a vertex in $G_2$). In this particular example $S$ contains all red edges of $G$ since each red edge connects a different (unordered) pair of connected components in $G_1$.