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.