Let us clarify these two issues first. The expansion of the exponential is: $$e^x = 1 + x + \mathcal{O}(x^2)\tag{1}\label{eq:exponential}$$ Second let us remind ourselves about integration by parts: $$\int \mathrm{d}^4\, x\; g(x)\square f(x) = \int \mathrm{d}^4\,x\; g(x)\partial_\mu\partial^\mu f = \sum_\mu g(x)\partial^\mu f(x)\bigg|_{\partial\mathbb{R}^4}-\int \mathrm{d}^4x\; \partial_\mu g(x)\partial^\mu f(x)\tag{2}\label{eq:intParts}$$ As it is common in this topic, fields are assume to vanish at infinity fast enough that boundary terms don't contribute. Now we can use \eqref{eq:intParts} two times, to place all the differential operators to act on $\phi$ instead of $\epsilon$, so that: $$-\frac{1}{2}\int\mathrm{d}^4 x\;\phi(x)\square\epsilon(x) = \frac{1}{2}\int\mathrm{d}^4x\;\partial_\mu\epsilon(x)\partial^\mu\phi(x) =-\frac{1}{2}\int\mathrm{d}^4x\;\epsilon(x)\square\phi(x).\tag{3}\label{eq:epsParts}$$

Starting with the third equation of the OP's question can use first equation \eqref{eq:exponential} to deal with the second factor: $$\langle \hat{\phi}(x)\rangle = \frac{1}{Z[0]}\int\mathcal{D}\phi e^{iS[\phi]}\left(1-\frac{i}{2}\int\mathrm{d}^4z\;(\phi(z)\square\epsilon(z)+\epsilon(z)\square\phi(z)) +\mathcal{O}(\epsilon^2)\right)\left[\phi(x) + \epsilon(x) \right] $$ where just the exponential having the $\epsilon$ field was expanded and $S$ stands for the kinetic term action of $\phi$. Now we can use the integration by parts as shown in \eqref{eq:epsParts}: $$\langle \hat{\phi}(x)\rangle = \frac{1}{Z[0]}\int\mathcal{D}\phi e^{iS[\phi]}\left(1-i\int\mathrm{d}^4z\;\epsilon(z)\square\phi(z)+\mathcal{O}(\epsilon^2)\right)\left[\phi(x) + \epsilon(x) \right]$$ Keeping only terms up to linear order in $\epsilon$ one obtains Schwartz's result: $$\langle \hat{\phi}(x)\rangle = \frac{1}{Z[0]}\int\mathcal{D}\phi e^{iS[\phi]}\left(\phi(x) + \epsilon(x) -i\phi(x)\int\mathrm{d}^4z\;\epsilon(z)\square\phi(z)\right)$$

Let us clarify these two issues first. The expansion of the exponential is: $$e^x = 1 + x + \mathcal{O}(x^2)\tag{1}\label{eq:exponential}$$ Second let us remind ourselves about integration by parts: $$\int \mathrm{d}^4\, x\; g(x)\square f(x) = \int \mathrm{d}^4\,x\; g(x)\partial_\mu\partial^\mu f = \int_{\partial\mathbb{R}^4} g(x)\partial^\mu f(x) \mathrm{d}\mathbf{n}_\mu -\int \mathrm{d}^4x\; \partial_\mu g(x)\partial^\mu f(x)\tag{2}\label{eq:intParts}$$ As it is common in this topic, fields are assume to vanish at infinity fast enough that boundary terms don't contribute, meaning the first term of the RHS is neglected. Now we can use \eqref{eq:intParts} two times, to place all the differential operators to act on $\phi$ instead of $\epsilon$, so that: $$-\frac{1}{2}\int\mathrm{d}^4 x\;\phi(x)\square\epsilon(x) = \frac{1}{2}\int\mathrm{d}^4x\;\partial_\mu\epsilon(x)\partial^\mu\phi(x) =-\frac{1}{2}\int\mathrm{d}^4x\;\epsilon(x)\square\phi(x).\tag{3}\label{eq:epsParts}$$

Starting with the third equation of the OP's question can use first equation \eqref{eq:exponential} to deal with the second factor: $$\langle \hat{\phi}(x)\rangle = \frac{1}{Z[0]}\int\mathcal{D}\phi e^{iS[\phi]}\left(1-\frac{i}{2}\int\mathrm{d}^4z\;(\phi(z)\square\epsilon(z)+\epsilon(z)\square\phi(z)) +\mathcal{O}(\epsilon^2)\right)\left[\phi(x) + \epsilon(x) \right] $$ where just the exponential having the $\epsilon$ field was expanded and $S$ stands for the kinetic term action of $\phi$. Now we can use the integration by parts as shown in \eqref{eq:epsParts}: $$\langle \hat{\phi}(x)\rangle = \frac{1}{Z[0]}\int\mathcal{D}\phi e^{iS[\phi]}\left(1-i\int\mathrm{d}^4z\;\epsilon(z)\square\phi(z)+\mathcal{O}(\epsilon^2)\right)\left[\phi(x) + \epsilon(x) \right]$$ Keeping only terms up to linear order in $\epsilon$ one obtains Schwartz's result: $$\langle \hat{\phi}(x)\rangle = \frac{1}{Z[0]}\int\mathcal{D}\phi e^{iS[\phi]}\left(\phi(x) + \epsilon(x) -i\phi(x)\int\mathrm{d}^4z\;\epsilon(z)\square\phi(z)\right)$$