The theorem as stated is true (see my comment though): If 1), 2), 3), and 4) hold, then $f_n$ converges uniformly to $f$.
The answer to your third question is "yes". This is the contrapositive of the theorem, which is logically equivalent to the theorem.
I think the answer to your second question is no (specifically, 4) does not necessarily have to hold).
Other observations:
Remark 1:
The theorem is "tight" (?); that is, each of the hypotheses are needed to insure that the convergence is uniform.
To see this:
Take $K=[0,1]$.
The functions $f_n(x)=x^n$ give a sequence satisfying 1), 3), and 4), but not 2).
$f_n$ converges to $f=0$ pointwise but does not converge uniformly to $f$ on $K$.
Take $K=[0,\infty)$ and $f_n(x)=\begin{cases} 0,\; & 0\leq x\leq n\\x-n , &n< x\leq n+1\\ 1 ,& x>n+1\end{cases}$. This sequence satisfies 1), 2) and 4) but not 3).
$f_n$ converges to $f=0$ pointwise but does not converge uniformly to $f$ on $K$.
Take $K=[0,1]$ and let $f_n(x)$ be the function whose graph consists of the straight line segment from $(0,0)$ to $({1\over2n},1)$, the straight line segment from $({1\over2n},1)$ to $({1\over n}, 0)$, and the straight line segment from $({1\over n},0)$ to $(1,0)$. Then $f_n$ satisfies 1), 2), and 3), but not 4).
$f_n$ converges to $f=0$ pointwise but does not converge uniformly to $f$ on $K$.
Take $K=[0,1]$ and $f_n(x)=\begin{cases}1, & 0\le x\le 1-{1\over n}\cr 0,&1-{1\over n}< x<1\\1,\;&x=1\end{cases} $. This sequence satisfies 2), 3) and 4) but not 1).
$f_n$ converges to $f=1$ pointwise but does not converge uniformly to $f$ on $K$.
Remark 2:
If $(f_n)$ is uniformly convergent to $f$, you may not conclude that all four conditions hold. For example $f_n(x)=\begin{cases}1/n,& 0\leq x<1/2\\ -1/n& 1/2\leq x\leq1 \end{cases}$ converges uniformly to the zero function. But $(f_n)$ does not satisfy 1) or 4).
This shows that the converse of the theorem is false. It is not an "if and only if" theorem.
I will expand this post later (if that's allowed).
I think your argument works, the only thing that isn't completely justified is why you can conclude that since $\hat{f}_n$ converges uniformly to $f$, so does $f_n$. And where do you use that there are only a finite number of points of discontinuity (the example below shows that this assumption is necessary). You can complete the argument as follows; Given $\epsilon>0$ there exists $\hat{N}$ and $\delta>0$ such that
$$ n>\hat{N} \ \Rightarrow \ \forall x \ : \ |\hat{f}_n(x)-f(x)|<\epsilon$$
Label the points of discontinuity $\{y_1,\ldots,y_k\}$. By pointwise convergence we know that for each point of discontinuity $y_j$, there exists and $N_j$ such that
$$ n>N_j \ \Rightarrow \ |f_n(x)-\hat{f}_n(x)|<\epsilon$$
But now taking $N=\max(\hat{N},N_1,\ldots,N_k)$ should give
$$ n>N \ \Rightarrow \ |f_n(x)-f(x)|<2\epsilon$$
EDIT:The following answers a misunderstood version of the question, but since reference to the example is made above, I will leave it up. Missed the assumption that points of discontinuity are assumed independent of $n$ and finite in number.
I am not convinced that this is necessarily true. How about the sequence of functions
$$ f_n(x) = \begin{cases} 0 & \text{ if } x\in ]-\infty,-1/n[\cup [0;\infty[\\
1/2+1/n & \text{ if }x\in [-1/n;0[ \\
\end{cases}$$
seems to me like this sequence of functions will converge pointwise to $0$ and be stictly decreasing, but it is clearly not uniformly convergent, since
$$ \forall n : \ \max(\{|f_n(-\delta) - f_n(0)| \ \big| \ \delta \in [-1;0]\}) >1/2$$
Best Answer
This is false. Let $a=0,b=1,f_n(x)=n-n^{2}x$ for $0<x<\frac 1 n$ and $f_n(x)=0$ for $x\geq \frac 1 n$. Can you check that this serves as a counterexample?