Why is $H_0(X,x_0) \ne H_0(X)$ 
We know $\tilde H_n(X) \cong \tilde H_n(X,x_0)=H_n(X,x_0)$ for all $n\ge 0$.
Also, when $n>0$, we have $H_n(X)\cong \tilde H_n(X) \cong H_n(X,x_0)$.
However, when $n=0$, we have $H_0(X)\cong \tilde H_0(X) \oplus \mathbb Z \cong H_0(X,x_0) \oplus \mathbb Z$. If $X$ is non-empty and path connected we have $H_0(X) \cong \mathbb Z$. So, $H_0(X,x_0) =0$.
So why do we have $H_0(X,x_0)=0$ but $H_0(X) \cong \mathbb Z$?
 A: As you said, for a pointed space $(X, x_0)$ the reduced homology is defined as 
$$\tilde{H}_n(X) = H(X, x_0) $$
Considering the inclusion of the basepoint $\iota\colon \{x_0\} \to X$ gives a long exact sequence
$$\dots \to H_n(x_0) \to H_n(X) \to H_n(X,x_0) \to H_{n-1}(x_0) \to\dots $$
Since $H_n(x_0)$ vanishes for $n>0$ we see that $H_n(X) \cong \tilde{H}_n(X)$ for all $n>0$. However we can also see that $H_0(X)\cong \tilde{H}_0(X)\oplus \mathbb{Z}$, so these don't agree in degree $0$. Intuitively, the $0$-th reduced homology group is ignoring the component of the basepoint. 
The intuition is that for pointed spaces we want to consider the information coming from the basepoint as being trivial. In singular homology this information is just a copy of the coefficient group in degree $0$ (indicating that a single point is connected) so our reduced homology ignores this group. Similarly, if $h$ is any homology (or cohomology) theory, then for a pointed space the inclusion of the basepoint induces a splitting
$$ h_*(X) \cong \tilde{h}_*(X)\oplus h_*(x_0) $$
This splitting is more interesting in theories like $K$-theory where $K^n(x_0)$ is non-zero for infinitely many values of $n$.
The reason we care about reduced homology is because sometimes it behaves more naturally than un-reduced, for example the suspension theorem $\tilde{H}_n(X) \cong \tilde{H}_{n+1}(\Sigma X)$ is not true for unreduced homology. Overall it's a more appropriate functor for the category of pointed spaces.
A: As you already seem to know that $H_0(X)\cong\mathbb Z$ for a path connected space $X$ and the standard argument for that is  geometric in nature  i will geometrically explain why $H_0(X,x_0)=0$.
Recall that  $H_n(X,A)$ is the homology of the chain complex 
$$...\to C_{n+1}(X)/C_{n+1}(A)\to C_n(X)/C_n(A)\to C_{n-1}(X)/C_{n-1}(A)\to ...$$
where $C_n(X)$ is a free abelian group with a basis given by all the $n$-simplices $\Delta^n\to X$ and boundary operator induced by the usual boundary operator. 
There is a more geometric description of this relative complex:
One can think of $C_n(X)/C_n(A)$ as a free abelian group with a basis given by all $n$-simplices $\Delta^n\to X$ whose image does not lie entirely in $A$ and with this interpretation applying the boundary operator is just applying the ususal boundary operator and then forgetting all simplices which lie entirely in $A$. We will now use this interpretation:
Let $X$ be a path connected space and $x_0\in X$. What are the cycles in $C_0(X)/C_0(\{x_0\})$? They are precisely elements generated by  the $0$-simplices $\sigma_x : \Delta^0\to \{x\}\subseteq X$ with $x\neq x_0$. However for any  such $x$ there is a path $\gamma$ from $x_0$ to $x$. Now the ususal boundary of the corresponding simplex $\sigma_\gamma$ is  $\sigma_x-\sigma_{x_0}$ but here  $\sigma_{x_0}$ gets ignored so the boundary is just $\sigma_x$. In particular $[\sigma_x]=0$ in $H_0(X,x_0)$, so $H_0(X,x_0)=0$.
