Hi stupid_question_bot,
Unfortunately you seem to need some more assumptions for an easy proof, in particular properness would make this very easy, in general given a flat proper scheme with geometrically normal fibers one can show that the number of (geometric) components of the fibers is locally constant on the base, which would answer your question.
(EDIT: To be clear, the following is not a counter-example to the specific statement in the question, that comes later in this answer. I was just trying to point out that the proof would need some geometric input since it is false when the base is not normal.)
The counter example that I have in mind is as follows, take $\bar{X}$ to be the nodal cubic over $\mathbb{Z}_p$ ($\mathbb{P}^1$ glued together at two $\mathbb{Z}_p$ points: say $0, 1$ in a standard affine chart), let $\bar{Y} \to \bar{X}$ be a connected finite etale cover corresopnding to a nontrivial element of the geometric fundamental group of $\bar{X}$ (for definiteness, take the double cover given by two $\mathbb{P}^1$'s glued into a bigon and for safety let $p \neq 2$). Now let $X$ be the complement of the node in the special fiber, and let $Y$ be the pullback. Clearly while the generic fiber of $Y$ is connected the special fiber is not by inspection.
You can now complain: "oh but your $X$ is not an snc complement in a smooth scheme." In this case I was unable to say anything useful, except for that some results in SGA imply that this would be true if the cover $Y$ is tamely ramified over the snc divisor. Hope this example is helpful though, as it shows that the strong statement you made about connectivity of special fibers is not some total triviality.
EDIT: Update, bad news: there are even worse examples to be had here. Let $X$ be $\mathbb{A}^1_{\mathbb{F}_p[[t]]}$, then consider $Y$ the Artin-Schreier cover of $X$ given by the equation $Y^p - Y = x \cdot t$, then the special fiber of this etale cover splits but generically it defines a Galois Artin-Schreier cover.
I think this could be a legitimate mistake. Namely, Milne does seem to implicitly be saying that the map $G(k^\mathrm{sep})\to \pi_0(G_{k^\mathrm{sep}})$ is a surjection. But, this does not have to be the case.
Example: Let $k$ be a separably closed, but not perfect field of characterististic $2$ (e.g. $k=\mathbb{F}_2(\!(t)\!)^\mathrm{sep}$). As $k$ is not perfect, there exists some $\alpha \in k$ which is not a square. Consider the group functor
$$G\colon \mathbf{Alg}_k\to \mathbf{Grp},\qquad A\mapsto \{a\in A: a^4=\alpha a^2\},$$
which one can consider as a group under addition. This is obviously a finite type group $k$-scheme represented by $\mathrm{Spec}(k[x]/(x^4-\alpha x^2)$. Let us then observe that
$$G_{\overline{k}}\cong \mathrm{Spec}(\overline{k}[x]/(x^4-\alpha x^2)\cong \mathrm{Spec}(\overline{k}[x]/((x^2-\beta x)^2),$$
where $\beta$ is a square root of $\alpha$ in $\overline{k}$. In particular, as thickenings by nil-ideals are homeomorphisms (see Tag 0BR6) we see that
$$|G_{\overline{k}}|\cong |\mathrm{Spec}(\overline{k}[x]/(x^2-\beta x))|\cong |\mathrm{Spec}(\overline{k}[x]/(x-\beta))|\sqcup |\mathrm{Spec}(\overline{k}[x]/(x)|\cong |\mathrm{Spec}(\overline{k})|\sqcup |\mathrm{Spec}(\overline{k})|,$$
(where $|\cdot|$ denotes the underlying topological space). In particular, we see that $\pi_0(G_{\overline{k}})$ has two elements.
The upshot here is that as $\overline{k}/k$ is a totally inseparable extension, the map $G_{\overline{k}}\to G$ is a homeomorphism (see Tag 0BRD) and so the map $\pi_0(G_{\overline{k}})\to \pi_0(G)$ is a bijection. So, $\pi_0(G)$ has two elements. That said, $G(k)$ has one element as if $a^4=\alpha a^2$ where $a$ is non-zero, then by dividing by $a$ we'd deduce that $\alpha$ is a square, which is a contradiction. Thus, the map $G(k)\to \pi_0(G)$ cannot be a surjection. $\blacksquare$
This is not really an issue though. If $G$ is an algebraic group over any field $k$, one can still define a group $k$-scheme structure on $\Pi_0(G)$ (where I am using $\Pi_0$ opposed to $\pi_0$ just to emphasize the group scheme versus the group: so $\Pi_0(G)(k'):=\pi_0(G_{k'})$) such that $\Pi_0(G)$ is finite etale and the map $G\to \Pi_0(G)$ is a group homomorphism (and, in fact, the initial one to etale group schemes). This just follows essentially by applying the functor $\Pi_0$ to the maps
$$m\colon G\times_k G\to G,\quad i\colon G\to G, \quad e\colon \mathrm{Spec}(k)\to G$$
and using the fact that $\Pi_0$ commutes with fiber products over $k$. This is explained in Chapter II, §5, № 1 of Demazure--Gabriel's book Groupes Algebriques, Tome I.
Of course, if $k$ is perfect, then what Milne says is a shortcut to this.
Best Answer
Let me record the discussion from the comments as an answer.
To compute the fiber over $P=(-1,0)$, take the spectrum of the tensor product $$k[s,t]/(t^2-(s^2-1)^2)\otimes_{k[x,y]/(y^2-x^3-x^2)} k$$ where $k$ is considered a $k[x,y]/(y^2-x^3-x^2)$-algebra as $(k[x,y]/(y^2-x^3-x^2))/(x+1,y)$. Since $M\otimes_R R/I\cong M/IM$, this tensor product is isomorphic to $k[s,t]/(t^2-(s^2-1)^2,s^2,st)$. Subtracting off multiples of $s^2$ from the first generator, we find that the ideal $(t^2-(s^2-1)^2,s^2,st)$ is equal to the ideal $(t^2-1,s^2,st)$. This means that $t^2=1$ in the quotient, so $1\cdot s =t^2\cdot s =0$, and therefore $s=0$. So the fiber over $P$ is the spectrum of $k[t]/(t^2-1)$, or two points, just like we want.
Added later: By the way, your claim that $f$ is flat by proposition III.9.7 is not correct - $Y$ is not regular, so this proposition doesn't apply. See here for more details about how to prove $f$ is flat.