This has been treated and answered already here if $G$ has a matching with $k$ edges, then it has a bipartite subgraph with $(|E(G)+k|)/2$ edges or more
I did not fully understand the probabilistic proof so therefore I am asking for clarification.
Let G have $m$ edges and a matching of size $\mu$ (i.e., $\mu$ edges with the property that no two of them share and endpoint (vertex)). Show that G has a bipartite subgraph with at least $\frac{\mu + m}{2}$ edges.
My attempt:
A bipartite subgraph can be constructed by first taking all the matchings edges and placing their vertices in two disjoint sets $V_1$ and $V_2$. We do that randomly and such that no edge in the matching is placed in $V_1$ or $V_2$. Thus the "randomness" is such that the vertices $v_i$ and $v_j$ of the matching-edge $(v_i,v_j)$ are placed with either $v_i$ in $V_1$ and $v_j$ in $V_2$ with probability $\frac{1}{2}$, or the other way around (also with probability $\frac{1}{2}$).
This procedure yields a bipartite subgraph with $\mu$ edges.
Now, if we take the remaining $m-\mu$ edges, those that were not in the matching, they have the property that each of their vertices is connected to at least one vertex in the matching. We put each of these vertices in $V_1$ with probability $\frac{1}{2}$ and in $V_2$ with probability $\frac{1}{2}$. For an arbitrary vertex among these vertices, we do not know how many vertices in the matching it is connected to. Let's say that $v_i$ is connected to $k_i$ vertices. Then the probability that all of these $k_i$ vertices are in the same set ($V_1$ or $V_2$) is (due to independence) $2\frac{1}{2^{k_i}} = 2^{1-k_i}$. This is the probability that the vertex does not contribute to a new edge in the bipartite subgraph.
Consider the random variable $I_{edge_i}$, the indicator that $v_i$ contributes with a new edge to the bipartite subgraph.
To get a lower bound on the number of edges, we can consider $\mu + \sum_i\text{E}\{I_{edge_i}\} = \mu + \sum_i\{1-2^{1-k_i}\}$
Problem: We don't know how many vertices $v_i$ like this there are, we do not know how many each of these vertices are connected to (the number $k_i$), so how can we calculate this last thing?
I confess I did not understand your attempt very well, but let me try to explain the statement that was made in the linked question:
Consider the matching $M$ with $\mu$ edges. For every edge $(v,w)$ of $M$, put $v$ in a set $V_1$ and $w$ in $V_2$ with probability $1/2$, and the other way around with probability $1/2$. That way we get two sets $V_1$ and $V_2$, each with $\mu$ vertices, such that two vertices that share an edge in $M$ are not in the same set.
Now flip a coin for each remaining vertex, such that $v$ is placed in $V_1$ with probability $1/2$ and in $V_2$ also with probability $1/2$.
The key observation is that if $v$ and $w$ do not share an edge in $M$, then the coin flips for $v$ and $w$ were independent (even if $v$ happens to share an edge of $M$ with some other vertex $u \neq w$). That is, $$Pr(v \in V_i \cap w \in V_j) = Pr(v \in V_i)Pr(w \in V_j) = \frac{1}{4}, i,j=1,2.$$
For an edge $(v,w)$ that is not in the matching (the graph is simple, so $v$ and $w$ do not share an edge in $M$), the probability that "its endpoints are in opposite sides" is $$Pr(v \in V_1 \cap w \in V_2) + Pr(v \in V_2 \cap w \in V_1) = \frac{1}{2} $$
Hope that clarifies the matter.