This answer attempts to give a simple algebraic intuition. Suppose $A$ is an $m \times n$ real matrix. Let $A=U\Sigma V^T$ be the SVD of $A$. Suppose that the rank of $A$ is equal to $r.$ Then the first $r$ singular values will be non-zero, while the remaining singular values will be zero.
If we write $U=[u_1 | \cdots | u_n]$ and $V=[v_1| \cdots | v_m]$, where $u_i$ is the $i^{th}$ column of $U$ (and similarly for $v_j$), then $A= \sum_{i=1}^r \sigma_i u_i v_i^T$, where $\sigma_i$ is the $i^{th}$ singular value. This shows that the linear transformation $A$ can be decomposed into the weighted sum of the linear transformations $u_i v_i^T$, each of which has rank $1$.
A large singular value $\sigma_k$ will indicate that the contribution of the corresponding transformation $u_k v_k^T$ is large and a small singular value will indicate that the corresponding contribution to the action of $A$ is small. As an application of this intuition, there are cases where e.g. $A$ is a full rank square matrix, hence it has no zero singular values, however a threshold is chosen and all terms in the sum $A= \sum_{i=1}^r \sigma_i u_i v_i^T$ corresponding to singular values less than this threshold are discarded. In that way, $A$ is approximated by a simpler matrix $\tilde{A}$, whose behavior is, for practical purposes, essentially the same as that of the original matrix.
It might also help to visualize the action of $A$ on a vector $x$ by means of the above formula: $Ax = \sum_{i=1}^r (\sigma_i\langle v_i,x\rangle) u_i $. Observe that the image of $x$ is a linear combination of the vectors $u_i$ and the coefficients depend on both the magnitude of the corresponding singular values as well as the directions of the vectors $v_i$ with respect to $x$. For example, if $x$ is orthogonal to all the $v_i$ for $i$ such that $\sigma_i \neq 0$, then $Ax=0$. On the other hand, if $x=v_k$ for some $k$ such that $\sigma_k \neq 0$, then $Av_k = \sigma_k u_k$.