Your second question has a negative answer because you've got some variances backwards. When $f:G \rightarrow G'$ is a map of groups, by composition with $f$ one gets a functor from the category of $G'$-sets to the category of $G$-sets. As Sawin notes, there is a functor from finite etale covers of $X$ to those of $Y$, namely base change, which computes the effect of composition with $\pi_1(Y,y) \rightarrow \pi_1(X,x)$ when $X$ and $Y$ are connected with respective geometric points $x$ and $y$ (over $x$).
Let's make this all totally down-to-earth by paying more attention to the role of the base points.
Let $k(y)$ and $k(x)$ denote the respective fields at $y$ and $x$, so there is a given $k$-embedding of $k(x)$ into $k(y)$. Choose $(X',x') \rightarrow (X,x)$ a Galois pointed connected finite etale cover with $k(x') = k(x)$, so $Y' := Y \times_X X'$ is a torsor over $X'$ for $\Gamma = {\rm{Aut}}(X'/X)$ (since $X'$ is a $\Gamma$-torsor over $X$, due to connectedness of $X'$ as a finite etale cover of $X$) that is equipped with a canonical $k(y)$-point $y'$ over $y$ via $k(y) \otimes_{k(x)} k(x') \simeq k(y)$. Each connected component of $Y'$ is therefore a Galois connected finite etale cover of $Y$, with Galois group given by its stabilizer in $\Gamma$. There is a preferred connected component $U(Y',y')$ of $Y'$, namely the one containing $y'$, and its covering group over $Y$ has just been seen to be a subgroup of $\Gamma$ in a natural way. Thus, we have a natural map
$$\pi_1(Y,y) \twoheadrightarrow {\rm{Aut}}(U(Y',y')/Y)^{\rm{opp}} \hookrightarrow
\Gamma^{\rm{opp}}.$$
This target is canonically a quotient of $\pi_1(X,x)$, and these maps compute the composition of $\pi_1(Y,y) \rightarrow \pi_1(X,x)$ by passage to the inverse limit over all $(X',x')$.
For your first question, you have to be more precise about the base points in order to formulate an answer. Consider a geometric point $x$ of $X$ over $k_s/k$ (i.e., there is a specified $k$-embedding of $k_s$ into the field at the geometric point $x$). Let $\kappa$ be the field at $x$. An element of $\pi_1(X,x)$ is a compatible system $\{f_{X',x'}:X' \simeq X'\}$ of $X$-automorphisms of a cofinal system of $\kappa$-pointed Galois connected finite etale covers $(X',x')$ of $X$ (but $f_{X',x'}$ certainly need not preserve $x'$!). For any finite Galois extension $K/k$ you would like to describe where the composite map $\pi_1(X,x) \rightarrow {\rm{Gal}}(k_s/k) \twoheadrightarrow {\rm{Gal}}(K/k)$ carries $\{f_{X',x'}\}$.
Consider the finite etale $X$-scheme $X_K$, which may or may not be connected. The field $\kappa$ contains $k_s$ over $k$ in a specified way, hence contains $K$ over $k$ in a specified way. Thus, there is a canonical map $K \otimes_k \kappa \rightarrow \kappa$, which is to say a $\kappa$-point $u(x,K)$ of $X_K$ over $x_K = {\rm{Spec}}(K \otimes_k \kappa)$. This point lies in a unique connected component $U(x,K)$ of $X_K$. But $X_K \rightarrow X$ is a ${\rm{Gal}}(K/k)$-torsor, so by degree considerations we see that each connected component is Galois over $X$ with covering group given by its stabilizer in ${\rm{Gal}}(K/k)$. In particular, $(U(x,K), u(x,K))$ is a $\kappa$-pointed Galois connected finite etale cover of $X$ with Galois group naturally inside ${\rm{Gal}}(K/k)$. Hence, $f_{U(x,K),u(x,K)}$ arises from a unique $\sigma(X,K) \in {\rm{Gal}}(K/k)$. This element is the image of $\{f_{X',x'}\}$ in ${\rm{Gal}}(K/k)$.
We see in particular by degree considerations (and counting sizes of various finite groups) that $\pi_1(X,x) \rightarrow {\rm{Gal}}(k_s/k)$ is surjective if and only if $U(x,K) = X_K$ for all $K$, which is to say $X$ is geometrically connected over $k$.
In the special case that $X$ is normal with function field $F = k(X)$ and with $X$ geometrically connected over $k$, we can make this even more explicit. Necessarily $k$ is separably closed in $F$ by the geometric connectedness hypothesis, so the separable closure $k_s$ of $k$ in a chosen separable closure $F_s$ of $F$ makes $k_s \otimes_k F$ a field naturally inside $F_s$ over $k$. This is visibly a Galois extension of $F$ with Galois group ${\rm{Gal}}(k_s/k)$. Hence, there is a natural quotient map
$${\rm{Gal}}(F_s/F) \twoheadrightarrow {\rm{Gal}}(k_s \otimes_k F/F) = {\rm{Gal}}(k_s/k).$$
Now use ${\rm{Spec}}(F_s) \rightarrow {\rm{Spec}}(F) \rightarrow X$ as the point $x$, so normality of $X$ implies that $\pi_1(X,x)$ is naturally a quotient of ${\rm{Gal}}(F_s/F)$. The above map of field Galois groups factors as the composition of ${\rm{Gal}}(F_s/F) \twoheadrightarrow \pi_1(X,x)$ with the canonical map $\pi_1(X,x) \rightarrow {\rm{Gal}}(k_s/k)$; in other words, for normal $X$ geometrically connected over $k$, the map is a refinement of the usual functoriality of absolutely Galois groups in the theory of fields (with ${\rm{Gal}}(k_s/k)$ playing the role of the opposite group of the connected pro-etale covering $X_{k_s} \rightarrow X$).
Best Answer
As Donu explained, the sequence splits by choosing an $\mathbf R$-point of $X$. So the only question remaining is what the $\operatorname{Gal}(\mathbf C/\mathbf R)$-action on $\pi_1^{\text{ét}}(X_{\mathbf C})$ is. I claim that the action is trivial, because the two points at infinity $V(x^2+y^2)$ are not defined over $\mathbf R$. (By contrast, for $\mathbf G_{m,\mathbf R}$ we get the nontrivial action where a generator $\sigma \in \operatorname{Gal}(\mathbf C/\mathbf R)$ acts by $-1$, using the same argument as below.)
Indeed, consider the tower $Y_n \to X_{\mathbf C} \to X$ where $Y_n \to X_{\mathbf C}$ is the unique cover of degree $n$. The composite $Y_n \to X$ is Galois as $\pi_1^{\text{ét}}(X_{\mathbf C}) \trianglelefteq \pi_1^{\text{ét}}(X)$ is normal and $n\mathbf Z \subseteq \mathbf Z$ is characteristic (in other words, any conjugate $Y'_n$ would contain $X_{\mathbf C}$ as $X_{\mathbf C} \to X$ is normal, hence $Y'_n \cong Y_n$ as $X_{\mathbf C}$-covers since $Y_n$ is the unique degree $n$ cover).
We need to compute $\operatorname{Gal}(Y_n/X)$. This depends on some choices of isomorphisms: we choose $X_{\mathbf C} \stackrel\sim\to \mathbf G_{m,\mathbf C}$ via $(x,y) \mapsto x+yi$, and we write $t = x+yi$. Under this identification, the generator $\sigma$ of $\operatorname{Gal}(X_{\mathbf C}/X) = \operatorname{Gal}(\mathbf C/\mathbf R)$ acts on $\mathbf G_{m,\mathbf C}$ via $x+yi \mapsto x-yi$, i.e. the $\mathbf C$-semilinear map $\sum_j c_jt^j \mapsto \sum_j \bar c_jt^{-j}$.
Then $Y_n$ can be identified with $\operatorname{Spec} \mathbf C[t^{\pm1/n}]$. The Galois group of $Y_n$ over $Y_1$ is $\mu_n(\mathbf C)$, where $\zeta \in \mu_n(\mathbf C)$ acts as the $\mathbf C$-linear map $t \mapsto \zeta t$. The conjugation action of $\sigma \in \operatorname{Gal}(X_{\mathbf C}/X)$ on $\operatorname{Gal}(Y_n/X_{\mathbf C})$ is therefore trivial, as $$(\sigma \circ \zeta \circ \sigma^{-1})(t) = (\sigma \circ \zeta)(t^{-1}) = \sigma (\zeta^{-1}t^{-1}) = \overline{\zeta^{-1}}t = \zeta t.$$ Thus, we see that the conjugation action of $\operatorname{Gal}(\mathbf C/\mathbf R)$ on $\pi_1^{\text{ét}}(X_{\mathbf C})$ is trivial. $\square$