We can fix $y$ and then consider $f(x,y)$ to be strictly a function of $x$ (say $h(x)=f(x,y)$ if you want, which makes sense because $y$ is fixed somewhere). Doing this yields the Taylor expansion seen in equation $(1)$. We would say that, by the usual formula, we have
$$h(x)=h(x_0)+h'(x_0)(x-x_0)+h''(x_0)(x-x_0)^2+\cdots \tag{0}$$
Now $h(x_0)=f(x_0,y)$ and $h'(x_0)=f_x(x_0,y)$ and $h''(x_0)=f_{xx}(x_0,y)$ and so forth. Now notice that this works regardless of what we fixed $y$ at to begin with, so it must hold for all available $y$.
We will switch gears now; since the formula holds for all $y$ we no longer need to consider $y$ fixed.
The Taylor expansion in the variable $x$ seen in $(0)$ involves a number of terms like $f_{xx}(x_0,y)$. Now we know that $x_0$ is fixed but $y$ is not, so this is a function of $y$! Say $g(y)=f_{xx}(x_0,y)$. Then we can speak of its Taylor expansion just as well, $g(y)=g(y_0)+g'(y)(y-y_0)+g''(y)(y-y_0)+\cdots$.
It turns out that $g(y_0)=f_{xx}(x_0,y_0)$ and $g'(y_0)=f_{xxy}(x_0,y_0)$ and $g''(y_0)=f_{xxyy}(x_0,y_0)$ and so on, because taking partial derivatives commutes with "evaluating at $y=y_0$" (or $x=x_0$, as in the last part). This means that differentiation and plugging things in can be done in any order here.
Doing a Taylor series expansion in the variable $y$ for $f_{xx}(x_0,y)$, as seen in $(4)$, can be done with any of the terms in $(1)$, like $f(x_0,y)$ and $f_x(x_0,y)$ seen in $(2)$ and $(3)$.
I have assumed that $f$ is sufficiently nice in this answer.
First make a table of your data. $$\begin {array}{r|r|r} i&x_i&f(x_i)\\ \hline 1&10&0\\2&10.2&0.004\\3&10.4&0.016\\4&10.6&0.036\\5&10.8&0.064\\6&11&0.1 \end {array}$$
Then, as the instructions say, $\phi_0=1$ and $\phi_1=x-\frac {(x\phi_0,\phi_0)}{\phi_0,\phi_0}=x-\frac {(x,1)}{(1,1)}$ The terms in parentheses are sums over the products at the points. So $(x,1)=10*1+10.2*1+\ldots 11*1=63, (1,1)=6$, which gives $\phi_1(x)=x-10.5$. Now you need to find $\phi_2$, which will be a quadratic.
Then you can express $f(x)=a_0\phi_0+a_1\phi_1+a_2\phi_2$ You can either find the $a$'s by expanding out the right side and equating like coefficients of $x$, or use the orthogonality to write $a_i=\frac {(f(x_i),\phi_i)}{(\phi_i,\phi_i)}$
Best Answer
If this is a fairly standard calculus course, you are working with the Lagrange form of the remainder. Notations tend to differ, but if $R_m(x)$ is the remainder when we truncate the Maclaurin series just after the $x^m$ term, then $$|R_m(x)|=\frac{|f^{(m+1)}(\xi)|}{(m+1)!} |x|^{m+1},$$ where $\xi$ is a number between $0$ and $x$.
In our case, all derivatives are $e^x$. If our $x$ is in the interval $[-2,2]$, then $|f^{(m+1)}(\xi)|\lt e^2$. And since $|x|\le 2$, we have $|x|^{m+1}\le 2^{m+1}$.
So in our situation, we have $$|R_m(x)|\lt \frac{e^2 2^{m+1}}{(m+1)!}.$$ Now you need to hunt around for an $m$ such that the above quantity is less than $1$. The expression is not a very nice one, so calculator experimentation is probably the best approach.
Remark: The problem is rather artificial, since in real work one would use the Maclaurin series for $x$ reasonably close to $0$, certainly not as far out as $x=2$.