First, you’ve reversed your conclusions: if the arguments were correct, the first would show that $\beta'(H)\le\alpha(H)$ and the second that $\beta'(H)\ge\alpha(H)$. The second argument is correct, but the first isn’t. To see where it goes wrong, apply it to $K_3$, a triangle. $X$ contains a single vertex, and both vertices in $Y$ are connected to that same vertex: your edge cover has cardinality $2$, not $1$ (and indeed $\alpha(K_3)=1$ while $\beta'(K_3)=2$).
Assume that $\alpha(H)=\beta'(H)$ for each subgraph $H$ of $G$ having no isolated vertices; you want to show that $G$ must be bipartite. HINT: If not, $G$ contains an odd cycle.
Edit, 11 Jue 2020
My original suggested proof for the other direction was nonsense, as pointed out by Mahsa in another answer while I was away and now by PAB in the comments; I tried to make it easier than it actually is. Here is a corrected version:
The result follows from three standard results.
- Let $\tau(G)$ be the vertex covering number of $G$; then $\alpha(G)+\tau(G)=|V(G)|$.
- Let $\nu(G)$ be the matching number of $G$; then $\nu(G)+\beta'(G)=|V(G)|$ if $G$ has no isolated vertex.
- (König) If $G$ is bipartite, then $\tau(G)=\nu(G)$.
Thus, if $G$ is bipartite and has no isolated points,
$$\beta'(G)=|V(G)|-\nu(G)=|V(G)|-\tau(G)=\alpha(G)\;.$$
The first two are fairly easy to prove.
For the first one, suppose that $X$ is a vertex cover such that $|X|=\tau(G)$. Then $V(G)\setminus X$ is an independent set of vertices, so $\alpha(G)\ge|V(G)|-\tau(G)$, i.e., $$\alpha(G)+\tau(G)\ge|V(G)|\;.$$ On the other hand, if $X$ is an independent set of vertices, and $|X|=\alpha(X)$, then $V(G)\setminus X$ is a vertex cover, so $\tau(G)\le|V(G)|-\alpha(G)$, i.e., $$\alpha(G)+\tau(G)\le|V(G)|\;.$$
For the second, let $F$ be an edge cover such that $|F|=\beta'(G)$. From each component of the graph $\langle V(G),F\rangle$ choose one edge; the result is a matching $M$. The graph $\langle V(G),F\rangle$ has at least $|V(G)|-|F|$ components — $\langle V(G),\varnothing\rangle$ has $|V(G)|$ components, and adding an edge reduces the number of components by at most $1$ — so $$\nu(G)\ge|M|\ge|V(G)|-|F|=|V(G)|-\beta'(G)\;,$$ i.e., $$\nu(G)+\beta'(G)\ge|V(G)|\;.$$
On the other hand, let $M$ be a matching such that $|M|=\nu(G)$. Let $W=V(G)\setminus V(M)$; since $M$ is maximal, $W$ is an independent set of vertices. There are no isolated vertices, so for each $w\in W$ we may choose an edge $e_w$ incident at $w$; let $F=\{e_w:w\in W\}$. Then $F\cup E(M)$ is an edge cover of $G$, so
$$\begin{align*}
\beta'(G)&\le|F|+|E(M)|=|V(G)|-|V(M)|+|E(M)|\\
&=|V(G)|-2\nu(M)+\nu(M)=|V(G)|-\nu(M)\;,
\end{align*}$$
i.e.,
$$\nu(G)+\beta'(G)\le|V(G)|\;.$$
For König’s theorem see here (and many other places).
The question is a little vague- for cut-edges, there is a precise answer, namely that the number of components increases by precisely one. This is fairly easy to see- if $e$ is a cut-edge, there will be one component of $G-e$ that contains one endpoint of $e$ and another component that contains the other end point. These are different components since $e$ is a cut-edge, and adding $e$ makes them into one component.
However, as you've already shown, you can't nail this down to an exact number for all cut-vertices, nor can it be determined just by the degree of the cut-vertex. I don't know of any generalized formula for how many components the deletion of a cut-vertex would add. I imagine the point of this question was for you to show that unlike cut-edges, cut-vertices do not add a predetermined number of components.
Best Answer
EDIT: Sorry - I did not think about threshold functions. I assumed we were working with a constant $p$, thought about the problem, and lost track of the actual question. I don't know how relevant it is, but I'll leave it.
Let $G$ be a bipartite graph on $2n$ vertices with parts $A$ and $B$ with $|A| = |B| = n$.
Let $E_A$ be the event that $A$ has an isolated vertex, $E_B$ be the event that $B$ has an isolated vertex. We're looking for $P(E_A \cup E_B) = P(E_A) + P(E_B) - P(E_A \cap E_B)$.
Each vertex in $A$ has $n$ possible edges, each with independent probability $p$, so the probability that it is isolated is $(1-p)^n$. Summing over the vertices of $A$, the probability that any vertex in $A$ is isolated is $n(1-p)^n$. The case for $B$ having an isolated is clearly symmetric. Hence, $P(E_A) = P(E_B) = n(1-p)^n$.
Now consider $P(E_A \cap E_B) = P(E_A)P(E_B|E_A)$. For any vertex in $B$, we have $n - 1$ possible edges (because we know one of the vertices in $A$ is isolated). The probability that a given vertex in $B$ is isolated is then $(1 - p)^{n - 1}$. Summing over the vertices of $B$, $P(E_B|E_A) = n(1-p)^{n-1}$. So $P(E_A \cap E_B) = (n(1-p)^n)(n(1-p)^{n-1}) = n^2(1-p)^{2n-1}$.
Putting it all together, $P(E_A \cup E_B) = 2n(1-p)^n + n^2(1-p)^{2n - 1}$.
As for a reference, I highly recommend Diestel's Graph Theory, which has an introductory treatment of random graphs (you may be above this level). There's a free preview here - http://diestel-graph-theory.com/. Bollobas' Modern Graph Theory has a similar but slightly deeper treatment. Bollobas also wrote a text called Random Graphs, which I know nothing about, but he is considered a leader in the field. I'm not sure that any of these address your specific problem.