Let $\mathcal{F}$ be a sheaf on a topological space $X$. Through Godement resolution $\mathcal{F}$ admits an exact sequence of sheaves $0\rightarrow\mathcal{F}\rightarrow\mathcal{F}^{0}\rightarrow\mathcal{F}^{1}\rightarrow\mathcal{F}^{2}\rightarrow\cdots$, where $\mathcal{F}^{i},i=0,1,\ldots$ are flasque. Define the $i$-th cohomology group $H^{i}(X,\mathcal{F})$ of a sheaf $\mathcal{F}$ on $X$ as follows:
Suppose $0\rightarrow\mathcal{F}\rightarrow\mathcal{F}^{0}\rightarrow\mathcal{F}^{1}\rightarrow\mathcal{F}^{2}\rightarrow\cdots$ is a flasque resolution of $\mathcal{F}$. Then $H^{i}(X,\mathcal{F})$ is the $i$-th cohomology group of the complex $\mathcal{F}^{0}(X)\rightarrow\mathcal{F}^{1}(X)\rightarrow\mathcal{F}^{2}(X)\rightarrow\cdots$. Explicitly, $H^{i}(X,\mathcal{F}):=\frac{Ker(\mathcal{F}^{i}(X)\rightarrow\mathcal{F}^{i+1}(X))}{Im(\mathcal{F}^{i-1}(X)\rightarrow\mathcal{F}^{i}(X))}$, $i\geq 1$ and $H^{0}(X,\mathcal{F}):=Ker(\mathcal{F}^{0}(X)\rightarrow\mathcal{F}^{1}(X))$.
To check the well-definedness we need to show that for any other flasque resolution $0\rightarrow\mathcal{F}\rightarrow\tilde{\mathcal{F}^{0}}\rightarrow\tilde{\mathcal{F}^{1}}\rightarrow\tilde{\mathcal{F}^{2}}\rightarrow\cdots$ of $\mathcal{F}$, the $i$-th cohomology group of $\mathcal{F}$ on $X$ remains the same modulo isomorphism.
I have showed the well-definedness of the $0$-th cohomology group as follows. Since $0\rightarrow\mathcal{F}\rightarrow\mathcal{F}^{0}\rightarrow\mathcal{F}^{1}\rightarrow\mathcal{F}^{2}\rightarrow\cdots$ and $0\rightarrow\mathcal{F}\rightarrow\tilde{\mathcal{F}^{0}}\rightarrow\tilde{\mathcal{F}^{1}}\rightarrow\tilde{\mathcal{F}^{2}}\rightarrow\cdots$ are exact sequence of sheaves and $\mathcal{F}\rightarrow\mathcal{F}^{0},\mathcal{F}\rightarrow\tilde{\mathcal{F}^{0}}$ are injective sheaf morphisms, by exactness, $Ker(\mathcal{F}^{0}(X)\rightarrow\mathcal{F}^{1}(X))=Im(\mathcal{F}(X)\rightarrow\mathcal{F}^{0}(X))\simeq\mathcal{F}(X)\simeq Im(\mathcal{F}(X)\rightarrow\tilde{\mathcal{F}^{0}}(X))=Ker(\tilde{\mathcal{F}^{0}}(X)\rightarrow\tilde{\mathcal{F}^{1}}(X))$
But I don't know how to establish the well-definedness of the higher cohomology groups. Any help is appreciated.
Proving that any old acyclic resolution computes sheaf cohomology is not too bad. Note that flasque sheaves are acyclic, so we can take care of your question with this general approach. First, let $0 \rightarrow \mathscr{F} \rightarrow \mathscr{A}^{\bullet}$ be an acyclic resolution. Because $\Gamma(X,-)$ is left exact, we know that $$ H^{0}(\Gamma(X,\mathscr{A}^{\bullet}))=\text{ker}\left(\Gamma(X,\mathscr{A}^{0}) \rightarrow \Gamma(X,\mathscr{A}^{1}) \right)=\Gamma(X,\mathscr{F})=H^{0}(X,\mathscr{F}).$$
Now consider the exact sequence
$$0 \longrightarrow \mathscr{F} \longrightarrow \mathscr{A}^{0} \longrightarrow K \longrightarrow 0.$$
From this, roll out the long exact sequence and observe we have isomorphisms $H^{i}(X,\mathscr{F}) \simeq H^{i-1}(X,K)$ for $i>1$, by the acyclicity of $\mathscr{A}^{0}$. However, $0 \rightarrow K \rightarrow \mathscr{A}^{1} \rightarrow \cdots$ is an injective resolution for $K$, so $H^{i-1}(X,K) \simeq H^{i}(\Gamma(X,\mathscr{A}^{\bullet}))$. In other words for all $i>1$ $$ H^{i}(X,\mathscr{F}) \simeq H^{i}(\Gamma(X,\mathscr{A}^{\bullet})).$$
Now for the case where $i=1$. We know that
$$H^{1}(\Gamma(X,\mathscr{A}^{\bullet})) \simeq \Gamma(X,K)/\text{im}\left(\Gamma(X,\mathscr{A}^{0}) \rightarrow \Gamma(X,\mathscr{A}^{1}) \right).$$
But because $K$ is the image of $\mathscr{A}^{0} \rightarrow \mathscr{A}^{1}$, we know that $$\text{im}\left(\Gamma(X,\mathscr{A}^{0}) \rightarrow \Gamma(X,\mathscr{A}^{1}) \right) \simeq \text{im}\left(\Gamma(X,\mathscr{A}^{0}) \rightarrow \Gamma(X,K) \right),$$
and thus, $H^{1}(X,\mathscr{F}) \simeq H^{1}(\Gamma(X,\mathscr{A}^{\bullet})$.