This is really just a matter of understanding the actual form of the excision isomorphism.
But we can pick $U$, $\sigma$, and $x$ somewhat carefully, to make this easier.
Instead of working with the given $\Delta$-complex structure, let's work with its 2nd barycentric subdivision, which is guaranteed to be an actual simplicial complex whose individual simplices are actually embedded. You probably know that a homology class is invariant under subdivision, so the two classes formed by summing the simplices of the 2nd barycentric subdivision and by summing the 2-simplices of the original $\Delta$-complex structure are equal. That gives us the freedom to work with teh 2nd barycentric subdivision.
Choose $\sigma$ to be a 2-simplex of the 2nd barycentric subdivision. And now choose $U$ to be a regular neighborhod of $\sigma$, chosen so small that $\sigma$ is the unique 2-simplex of the 2nd barycentric subdivision that is contained in $U$; because it's a regular neighborhood of a closed disc in a manifold, $U$ is homeomorphic to $\mathbb R^2$.
Finally, pick $x$ to be a point in the interior of $\sigma$.
What we need to show is that if $c$ is the sum of all simplices of the 2nd barycentric subdivision then the class $[c] \in H_2(M,M-x)$ is the equal to the image, under excision, of the class $[\sigma] \in H_2(U,U-x)$. And this is really just a matter of understanding a concrete description of the excision homomorphism.
Excision can be described like this:
If you have a $k$-cycle $c = \sum a_i \tau_i$ of $(M,M-x)$, and if each $\tau_i$ is contained in either $M-x$ or $U$, and if $c'$ is obtained from $c$ by discarding all terms $a_i\tau_i$ such that $\tau_i$ is contained in $M-x$, then $c'$ is a $k$-cycle of $(U,U-x)$, and the excision map $H_k(U,U-x) \to H_k(M,M-x)$ takes $[c']$ to $[c]$.
Now let's apply this. Certainly the given $c$ is a cycle of $M$, and it is in fact a fundamental cycle, representing the fundamental class $[M]$. So $c$ is certainly also a cycle of $(M,M-x)$. Applying the above description of excision, the terms that we remove from $c$ are all of the terms except for $\sigma$, which is the only term contained in $U$. So all that's left is $c'=\sigma$, and we're done.