Why is the Monotone Convergence Theorem restricted to a nonnegative function sequence?

Well, if $f_k$ could be negative, then its integral might not even be defined. For instance, if $X=\mathbb{R}$ with Lebesgue measure and $f_k(x)=x$ for some $k$, there is no good way to define $\int f_k$ (it should morally be "$\infty-\infty$"). On the other hand, the integral of a nonnegative measurable function can always be defined (though it might be $\infty$).

Even if you require $\int f_k$ to be defined for all $k$, if $\int f_k$ is allowed to be $-\infty$, the result can be false. For instance, let $X=\mathbb{N}$ with counting measure and let $f_k(n)=-1$ if $n>k$ and $0$ if $n\leq k$. Then the $f_k$ are monotone increasing and converge pointwise to the constant function $0$, but $\int f_k=-\infty$ for all $k$.

On the other hand, if you require $\int f_k$ to be defined and $>-\infty$ for all $k$, the result is true. Indeed, you can just replace each $f_k$ by $f_k-f_1$ and use the usual version of the theorem, since all these functions are nonnegative (and the equation $\int f_k=\int f_1+\int (f_k-f_1)$ is guaranteed to make sense and be true since $\int f_1>-\infty$).

If they could be negative, then the statement would also have to be true for any sequence $f_n$ where $f_n(x)\geq f_{n+1}(x)$ by just reflecting across $0$. But then consider when $f_n:[0,\infty]\rightarrow R, f_n(x)=\frac{x}{n}$.

In this case, for each $x$, $\lim_n f_n(x)=0$, and thus $f_n \rightarrow f=0$ pointwise, but the integrals are infinite for each finite $n$.