Related to question (1), suppose you wanted to get a 4-manifold with boundary as a submanifold of $\mathbb{R}^4$ with fundamental group a finitely presented group, with presentation complex $K$. It is a result of Curtis that any 2-complex $K$ is homotopy equivalent to a 2-complex which embeds in $\mathbb{R}^4$ (see also the Stallings reference in the comment on this MO question and Dranishnikov-Repovs - in fact this works if the 2-complex $K$ has only one vertex). However, to thicken up $K\subset \mathbb{R}^4$ to get a tubular neighborhood which is a 4-manifold (with boundary) in $\mathbb{R}^4$ which retracts to $K$, $K$ must be nicely embedded. For example, the 2-cells should be locally flat. I'm not sure if this holds true for the embeddings of Curtis or for the other constructions.

To answer (2), think about what happens when you thicken up the complex. Thickening some points in $\mathbb{R}^5$ gives some 5-balls, whose boundary is a union of 4-spheres. Thickening an interval attached to some points, one gets a 1-handle $D^1\times D^4$, whose boundary removes two 4-balls from the 4-spheres and attaches in $D^1 \times S^3$, giving a 4-manifold with free fundamental group (a connect sum of $S^3\times S^1$'s). The 2-cells thicken up to 2-handles $D^2\times D^3$, which remove $S^1\times D^3$ from the 4-manifold (which doesn't change the fundamental group, since $S^1$ is codimension 3), and replaces it with $D^2 \times S^2$, which has the effect of killing the element in the free group corresponding to the circle by Van Kampen. So you see that this gives the same thing as the construction in Henry Wilton's answer to the other question.

Related to question (1), suppose you wanted to get a 4-manifold with boundary as a submanifold of $\mathbb{R}^4$ with fundamental group a finitely presented group, with presentation complex $K$. It is a result of Curtis that any 2-complex $K$ is homotopy equivalent to a 2-complex which embeds in $\mathbb{R}^4$ (see also the Stallings reference in the comment on this MO question and Dranishnikov-Repovs - in fact this works if the 2-complex $K$ has only one vertex). However, to thicken up $K\subset \mathbb{R}^4$ to get a tubular neighborhood which is a 4-manifold (with boundary) in $\mathbb{R}^4$ which retracts to $K$, $K$ must be nicely embedded. For example, the 2-cells should be locally flat. I'm not sure if this holds true for the embeddings of Curtis or for the other constructions.

To answer (2), think about what happens when you thicken up the complex. Thickening some points in $\mathbb{R}^5$ gives some 5-balls, whose boundary is a union of 4-spheres. Thickening an interval attached to some points, one gets a 1-handle $D^1\times D^4$, whose boundary removes two 4-balls from the 4-spheres and attaches in $D^1 \times S^3$, giving a 4-manifold with free fundamental group (a connect sum of $S^3\times S^1$'s). The 2-cells thicken up to 2-handles $D^2\times D^3$, which remove $S^1\times D^3$ from the 4-manifold (which doesn't change the fundamental group, since $S^1$ is codimension 3), and replaces it with $D^2 \times S^2$, which has the effect of killing the element in the free group corresponding to the circle by Van Kampen. So you see that this gives the same thing as the construction in Henry Wilton's answer to the other question.

Related to question (1), it is a result of Curtis that any 2-complex $K$ is homotopy equivalent to a 2-complex which embeds in $\mathbb{R}^4$ (see also the Stallings reference in the comment on this MO question and Dranishnikov-Repovs - in fact this works if the 2-complex $K$ has only one vertex). However, to thicken up $K\subset \mathbb{R}^4$ to get a tubular neighborhood which is a 4-manifold in $\mathbb{R}^4$ which retracts to $K$, $K$ must be nicely embedded. For example, the 2-cells should be locally flat. I'm not sure if this holds true for the embeddings of Curtis or for the other constructions.

To answer (2), think about what happens when you thicken up the complex. Thickening some points in $\mathbb{R}^5$ gives some 5-balls, whose boundary is a union of 4-spheres. Thickening an interval attached to some points, one gets a 1-handle $D^1\times D^4$, whose boundary removes two 4-balls from the 4-spheres and attaches in $D^1 \times S^3$, giving a 4-manifold with free fundamental group (a connect sum of $S^3\times S^1$'s). The 2-cells thicken up to 2-handles $D^2\times D^3$, which remove $S^1\times D^3$ from the 4-manifold (which doesn't change the fundamental group, since $S^1$ is codimension 3), and replaces it with $D^2 \times S^2$, which has the effect of killing the element in the free group corresponding to the circle by Van Kampen. So you see that this gives the same thing as the construction in Henry Wilton's answer to the other question.

Related to question (1), suppose you wanted to get a 4-manifold with boundary as a submanifold of $\mathbb{R}^4$ with fundamental group a finitely presented group, with presentation complex $K$. It is a result of Curtis that any 2-complex $K$ is homotopy equivalent to a 2-complex which embeds in $\mathbb{R}^4$ (see also the Stallings reference in the comment on this MO question and Dranishnikov-Repovs - in fact this works if the 2-complex $K$ has only one vertex). However, to thicken up $K\subset \mathbb{R}^4$ to get a tubular neighborhood which is a 4-manifold (with boundary) in $\mathbb{R}^4$ which retracts to $K$, $K$ must be nicely embedded. For example, the 2-cells should be locally flat. I'm not sure if this holds true for the embeddings of Curtis or for the other constructions.

To answer (2), think about what happens when you thicken up the complex. Thickening some points in $\mathbb{R}^5$ gives some 5-balls, whose boundary is a union of 4-spheres. Thickening an interval attached to some points, one gets a 1-handle $D^1\times D^4$, whose boundary removes two 4-balls from the 4-spheres and attaches in $D^1 \times S^3$, giving a 4-manifold with free fundamental group (a connect sum of $S^3\times S^1$'s). The 2-cells thicken up to 2-handles $D^2\times D^3$, which remove $S^1\times D^3$ from the 4-manifold (which doesn't change the fundamental group, since $S^1$ is codimension 3), and replaces it with $D^2 \times S^2$, which has the effect of killing the element in the free group corresponding to the circle by Van Kampen. So you see that this gives the same thing as the construction in Henry Wilton's answer to the other question.