Anti - De Sitter Space is the maximally symmetric solution to field equations with negative cosmological constant.
The negative cosmological constant also shows that the spacetime has negative curvature. I don't understand how is the cosmological constant related to the curvature of spacetime? Also how do we show this mathematically that Anti-De Sitter spacetime has negative curvature?
The local radius of curvature $r(x)$ at a point $x$ in a spacetime is effectively given by the formula $$ R(x) = \frac{1}{r(x)^2} $$ where $R$ Ricci scalar.
AdS space is a maximally symmetric space satisfying Einstein's equations with a negative cosmological constant, $$ R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R + \Lambda g_{\mu\nu} = 0 \, , \qquad \Lambda < 0 \,. $$ Then, taking a trace of the equation above, we find $$ R - \frac{1}{2} d R + \Lambda d = 0 ~~\implies ~~ R = \frac{2\Lambda d}{d-2} $$ Thus, $$ r^2 = \frac{d-2}{2\Lambda d} \, . $$ Here, $d$ is the dimension of the AdS space. For $d>2$, the RHS of the above equation is negative (since $\Lambda < 0$). Thus, the spacetime has constant (because $\Lambda$ is a constant) negative curvature.
Be careful about trying to visualise the curvature in a rubber sheet fashion. Anti-de Sitter space is a Lorentzian manifold not a Riemannian manifold and this messes up our attempts at an intuitive image. There is a useful discussion of a related issue in Why does dark energy produce positive space-time curvature?
When we say AdS space is negatively curved this has a precise technical meaning i.e. the scalar curvature is negative. The scalar curvature is obtained by contracting the Ricci tensor, which is in turn obtained from the Riemann curvature tensor for AdS spacetime.
The accepted answer to the question I've linked shows a simple way to relate the sign of the scalar curvature to the sign of the cosmological constant using the Einstein equation, but I'm not sure I would say this is intuitive. I don't know of any simple intuitive way to relate the two.
Footnote: I see Prahar's answer explains the calculation I was referring to i.e. starting with the Einstein equation and taking the trace to relate $R$ and $\Lambda$. I'll leave you to decide if this counts as intuitive.
The answer to OP's questions is a result of the following calculations:
Calculate the curvature tensor from the AdS metric.
Next determine the cosmological constant from the matter-free EFE.
