Assume one has a Hilbert space of dimension ${K= mn}$. It is possible to describe it in (at least) two ways — simply as a $K$-dimensional space, or as a product of $m$- and $n$-dimensional spaces. In the former case we shall call it $\mathcal{H}$, in the latter — $\widetilde{\mathcal{H}}$. All the states in $\mathcal{H}$ and ${\widetilde{\mathcal{H}}=\widetilde{\mathcal{H}}_1\otimes \widetilde{\mathcal{H}}_2}$, as well as the quantum operations, should be in one-to-one correspondence with each other. So, mathematically these two descriptions are equivalent. Let us now explore whether anything is changing from the physical point of view.
For example, in the case of $\widetilde{\mathcal{H}}$ we can measure only a `part' of the state corresponding to one of the Hilbert spaces $\widetilde{\mathcal{H}}_1$ and $\widetilde{\mathcal{H}}_2$. This will cause the partial collapse of the wave function. By performing measurements in $\widetilde{\mathcal{H}}_1$, one gets (here and in what follows for simplicity we assume ${\dim \widetilde{\mathcal{H}}_1 = \dim \widetilde{\mathcal{H}}_2 = 2}$, and also that the measurements in $\widetilde{\mathcal{H}}_{1,2}$ are performed in the computational basis):
\begin{alignat}{9}
\label{pr1}
\dfrac{1}{2}(|0\rangle+|1\rangle) \otimes (|0\rangle+|1\rangle) &\longrightarrow
\left[\begin{alignedat}{9}
&|0\rangle\otimes \dfrac{1}{\sqrt{2}} (|0\rangle+|1\rangle)\\
&|1\rangle\otimes \dfrac{1}{\sqrt{2}} (|0\rangle+|1\rangle)
\end{alignedat}\right.\quad&&,\qquad\qquad(1)\\
\label{pr2}
\dfrac{1}{\sqrt{2}}
(|0\rangle\otimes|0\rangle+|1\rangle\otimes|1\rangle)
&\longrightarrow
\left[\begin{alignedat}{9}
&|0\rangle\otimes |0\rangle \\
&|1\rangle\otimes |1\rangle
\end{alignedat}\right.\quad&&.\qquad\qquad(2)
\end{alignat}
In the former case, the measurement in $\widetilde{\mathcal{H}}_1$ does not affect the part of the state belonging to $\widetilde{\mathcal{H}}_2$. In the latter case the situation is opposite — due to entanglement, the outcome of the $\widetilde{\mathcal{H}}_1$ measurement ensues the collapse of $\widetilde{\mathcal{H}}_2$. It natural to ask what are the reflections of these effects in the case of $\mathcal{H}$.
Denoting the four states of $\mathcal{H}$ with letters, we assign the states as:
\begin{alignat}{99}
|00\rangle & = |a\rangle \quad&&,\qquad
|01\rangle & = |b\rangle \quad&&,\qquad
|10\rangle & = |c\rangle \quad&&,\qquad
|11\rangle & = |d\rangle \quad&&.\qquad\qquad(3)
\end{alignat}
We now want to recast (1) and (2) in terms of $\mathcal{H}$:
\begin{alignat}{9}
\label{pr3}
\dfrac{1}{2}(|a\rangle+|b\rangle+|c\rangle+|d\rangle) &\longrightarrow
\left[\begin{alignedat}{9}
&\dfrac{1}{\sqrt{2}} (|a\rangle+|b\rangle) \\
&\dfrac{1}{\sqrt{2}} (|c\rangle+|d\rangle)
\end{alignedat}\right.\quad&&,\qquad\qquad(4)\\
\label{pr4}
\dfrac{1}{\sqrt{2}}
(|a\rangle+|d\rangle)
&\longrightarrow
\left[\begin{alignedat}{9}
&|a\rangle \\
&|d\rangle
\end{alignedat}\right.\quad&&.\qquad\qquad(5)
\end{alignat}
It's tempting to call the operations above `projecting onto the states on the RHS of (1)'. To convince ourselves, let us take a look at the quantum channels corresponding to these measurements. The $\widetilde{\mathcal{H}}_1$ measurement operators of $\widetilde{\mathcal{H}}$ are:
\begin{equation}
\widetilde{M}_1 = \dfrac{1}{\sqrt{2}} (|0\rangle\langle0|) \otimes {\rm 1\hspace{-0.4ex}\rule{0.1ex}{1.52ex}\hspace{0.2ex}}
\qquad,\quad
\widetilde{M}_2 = \dfrac{1}{\sqrt{2}} (|1\rangle\langle1|) \otimes {\rm 1\hspace{-0.4ex}\rule{0.1ex}{1.52ex}\hspace{0.2ex}}
\quad.\qquad\qquad(6)
\end{equation}
They represent a complete set of measurement operators in $\widetilde{\mathcal{H}}_1$, and, of course, an incomplete set in $\widetilde{\mathcal{H}}$. Rewriting those in the basis of $\mathcal{H}$ renders:
\begin{equation}\begin{alignedat}{9}
M_1 &= \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \otimes {\rm 1\hspace{-0.4ex}\rule{0.1ex}{1.52ex}\hspace{0.2ex}}
= \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\\end{pmatrix} = |a\rangle\langle a| + |b\rangle\langle b| \quad&&,\qquad\qquad(7)\\
M_2 &= \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \otimes {\rm 1\hspace{-0.4ex}\rule{0.1ex}{1.52ex}\hspace{0.2ex}}
= \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1\\\end{pmatrix} = |c\rangle\langle c| + |d\rangle\langle d| \quad&&.\qquad\qquad(8)
\end{alignedat}\end{equation}
As expected, these are the projectors onto $\operatorname{span}\{|a\rangle,|b\rangle\}$ and $\operatorname{span}\{|c\rangle,|d\rangle\}$.
Now, speaking of physics. What would one say about the state ${\dfrac{1}{2}(|0\rangle+|1\rangle) \otimes (|0\rangle+|1\rangle)}$ when measuring in the computational basis? — 'Not sure about either of two qubits'. And about the post-measurement state ${|0\rangle\otimes \dfrac{1}{\sqrt{2}} (|0\rangle+|1\rangle)}$? — 'Not sure about the second qubit but the first one is definitely $|0\rangle$'. Clearly, (4) permits for the similar interpretation.
The celebrated feature of the maximally entangled Bell state is the observer's ability to completely identify the post-measurement state of a composite system by performing a measurement on its part. By looking at (5), we see how this mechanism works in the language of $\mathcal{H}$: the identification of a general post-measurement state after projecting the input on a two-dimensional subspace is, of course, impossible. However, for certain special states such a `soft' measurement may give the full information about the post-measurement state.