Here is a counterexample for surjectivity: $$B=k[x,y], A=B[X,Y]/(xX+yY-1)$$ It is easy to see that the image of the induced morphism is the plane minus the origin.

It is not hard neither to see that the extension is algebraically closed: If $a\in \text{Frac}(A)=k(x,y,X)$ is algebraic over $\text{Frac}(B)=k(x,y)$, then it must be in $k(x,y)$ as this is a purely transcendental field extension. As $k[x,y]$ is integrally closed, $a$ must actually be in $k[x,y]$.

Here is a counterexample for surjectivity: $$B=k[x,y], A=B[X,Y]/(xX+yY-1)$$ It is easy to see that the image of the induced morphism is the plane minus the origin.

Now suppose $a\in A$ is a root of a nonzero polynomial with coefficients in $B$. As $\text{Frac}(A)=k(x,y,X)$ is a purely transcendental extension of $\text{Frac}(B)=k(x,y)$, then $a$ must be in $\text{Frac}(B)$. Write $a=\frac{p}{q}$ for $p,q\in B$ relatively prime. If $a\not\in B$ then $q$ is not a unit, so its zero locus has dimension $1$, so it contains a point $(x_0,y_0)$ other than the origin. Then we can choose $X_0,Y_0$ so that $x_0X_0+y_0Y_0=1$, and then $a$ has a pole at $(x_0,y_0,X_0,Y_0)$, which is a contradiction as $a\in A$.