The formulations are equivalent.
By transposing $X$ if necessary, we may reduce the situation to where $X$ has at least as many rows, $n$, as columns, $p$. Consider the decomposition of $X$ into
$$X = U\Sigma V^\prime$$
for an $n\times n$ orthogonal matrix $U$, an $n\times p$ matrix $\Sigma$ that is diagonal in the sense that $\Sigma_{ij}=0$ whenever $i\ne j$, and a $p\times p$ orthogonal matrix $V$. This $\Sigma$ can be considered to be a diagonal $p\times p$ matrix $S$ stacked on top of a $(n-p)\times p$ matrix of zeros, $Z$. The effect of $Z$ in the product $U\Sigma$ is to "kill" the last $n-p$ columns of $U$. We may therefore drop those columns and drop $Z$, producing a decomposition
$$X = U_0 S V^\prime$$
where the columns of $U_0$--being the first $p$ columns of $U$--are orthogonal. The dimensions of these matrices are $n\times p$, $p\times p$, and $p\times p$.
Conversely--there's a theorem involved here--we may always extend an $n\times p$ matrix $U_0$ of orthogonal (and unit length) columns into an orthogonal $n\times n$ matrix. Geometrically this is obvious--you can always complete a partial basis of $p$ unit length, mutually perpendicular vectors into a full basis of such vectors--and algebraically it is performed using the Gram-Schmidt process. ($U$ is not unique unless $n=p$.) After doing this, stack an $n-p\times p$ matrix of zeros underneath $S$ to produce $\Sigma$ and you have recovered the first decomposition.
A more modern conception of matrices and their multiplication is that any $a\times b$ matrix $Y$ represents a linear transformation $T_Y:V^{(b)}\to V^{(a)}$ where $V^{(a)}$ is a vector space of dimension $a$ and $V^{(b)}$ a vector space of dimension $b$, and multiplication represents functional composition. The matrix $V^\prime$ maps $\mathbb{R}^p$ to itself, $\Sigma$ maps $\mathbb{R}^p$ into the subspace of $\mathbb{R}^n$ spanned by the first $p$ components, and $U$ maps $\mathbb{R}^n$ into itself. Since the image of $V^\prime$ is only this $p$-dimensional subspace, $U$ and its restriction to the subspace (called $U_0$ above) agree on the image of $\Sigma$. Moreover, $\Sigma$ decomposes as a linear transformation $S:\mathbb{R}^p\to\mathbb{R}^p$ followed by a fixed embedding of $\mathbb{R}^p$ into $\mathbb{R}^n$. (The matrix of this embedding is a diagonal $n\times p$ matrix with all ones on the diagonal.) This makes it geometrically obvious that the transformations from $\mathbb{R}^p$ to $\mathbb{R}^n$ represented by $U_0 S V^\prime$ and $U\Sigma V^\prime$ are the same.
