This answer tries to give more connections between these two decompositions than their differences.
SVD actually stems from the eigenvalue decomposition of real symmetric matrices. If a matrix $A \in \mathbb{R}^{n \times n}$ is symmetric, then there exists an real orthogonal matrix $O$ such that
$$A = O\text{diag}(\lambda_1, \ldots, \lambda_n)O', \tag{1}$$
where $\lambda_1, \ldots, \lambda_n$ are all real eigenvalues of $A$. In other words, $A$ is orthogonal similar to a diagonal matrix $\text{diag}(\lambda_1, \ldots, \lambda_n)$.
For a general (rectangular) real matrix $B \in \mathbb{R}^{m \times n}$, clearly $B'B$ is square, symmetric and semi-positive definite, thus all its eigenvalues are real and non-negative. By definition, the singular values of $B$ are all arithmetic square root of the positive eigenvalues of $B'B$, say, $\mu_1, \ldots, \mu_r$. Since $B'B$ has its eigen-decomposition
$$B'B = O\text{diag}(\mu_1^2, \ldots, \mu_r^2, 0, \ldots, 0)O',$$
it can be shown (doing a little clever algebra) that there exist orthogonal matrices $O_1 \in \mathbb{R}^{m \times m}$ and $O_2 \in \mathbb{R}^{n \times n}$ such that $B$ has the following Singular Value Decomposition (SVD):
$$B = O_1 \text{diag}(\text{diag}(\mu_1, \ldots, \mu_r), 0)O_2, \tag{2}$$
where $0$ in the diagonal matrix is a zero matrix of size $(m - r) \times (n - r)$. $(2)$ sometimes is said as $B$ is orthogonal equivalent to the diagonal matrix $\text{diag}(\text{diag}(\mu_1, \ldots, \mu_r), 0)$.
In view of $(1)$ and $(2)$, both eigen-decomposition (in its narrow sense for symmetric matrices only) and SVD are trying to look for representative elements under some relations.
In detail, the eigen-decomposition $(1)$ states that under the orthogonal similar relation, all symmetric matrices can be classified into different equivalent classes, and for each equivalent class, the representative element can be chosen to be the simple diagonal matrix $\text{diag}(\lambda_1, \ldots, \lambda_n)$. It can be further shown that the set of eigenvalues $\{\lambda_1, \ldots, \lambda_n\}$ is the maximal invariant under the orthogonal similar relation.
By comparison, the SVD $(2)$ states that under the orthogonal equivalent relation, all $m \times n$ matrices can be classified into different equivalent classes, and for each equivalent class, the representative element can also be chosen to be a diagonal matrix $\text{diag}(\text{diag}(\mu_1, \ldots, \mu_r), 0)$. It can be further shown that the set of singular values $\{\mu_1, \ldots, \mu_r\}$ is the maximal invariant under the orthogonal equivalent relation.
In summary, given a matrix $M$ to be decomposed, both eigen-decomposition and SVD aim to seek for its simplified profile. This is not much different from seeking a representative basis under which a linear transformation has its simplistic coordinate expression. Moreover, the above (incomplete) arguments showed that eigen-decomposition and SVD are closely related -- in fact, one way to derive SVD is completely from the eigen-decomposition.