You should look at this in terms of energy. (We only consider the perpendicular velocity component as the ball hits the wall.)
The ball has kinetic energy $K=½mv^2$. As it hits the wall in a fully elastic collision (the wall doesn't move), all the kinetic energy has to be kept inside the ball. In real situations some energy will turn into heat in the ball and wall - but in a simple model we see all the energy being reconverted into kinetic energy in the ball again.
(If you do have an angle (the V-shape), then you may also consider potential gravitational energy, but that will only have an impact on the y-component of the velocity and the kinetic energy calculated from that.)
When the ball hits the wall, the wall does work on the ball:
The work done is the double kinetic energy.
You are right that the force that the wall exerts on the ball is equal. Look at it in two stages.
The ball is decelerated to zero speed (all kinetic energy i stored as elastic potential energy in the wall/ball). The happens because the all hits the wall with force $F_1$, and the wall the reacts with force $-F_1$ to stop the ball.
The ball is now (in this intermediate position with zero speed) a bit compressed. As it uncompresses as a spring, it will now exert a force on the wall to push itself away, $F_2$. The wall pushes back with $-F_2$. The point is now that the force it took to compress the ball (to stop it) is the same force as the compressed ball uses to uncompress (like a spring). So $F_1=F_2$, and thus the acceleration that happens is the same as the deceleration that stopped the ball first, just with another sign.
And then the spring-back effect from this collision happens.