(this is pretty much what Michael said, I just happen to have a few notes on this and I thought this might be helpful. If I were you I would try to extrapolate from Lagrange's form of the remainder)
We already know Taylor's theorem for functions on $\mathbb{R}$, $ g(x) = g(a)+g'(a)(x-a) + \frac{1}{2}g''(a)(x-a)^2 +\cdots +\frac{1}{k!}g^{(k)}(a)(x-a)^k + R_k $ and... If the remainder term vanishes as $k \rightarrow \infty$ then the function $g$ is represented by the Taylor series given above and we write: $ g(x) = \sum_{k=0}^{\infty}\frac{1}{k!}g^{(k)}(a)(x-a)^k. $ Consider the function of two variables $f: U \subseteq \mathbb{R}^2 \rightarrow \mathbb{R}$ which is smooth with smooth partial derivatives of all orders. Furthermore, let $(a,b) \in U$ and construct a line through $(a,b)$ with direction vector $(h_1,h_2)$ as usual: $ \phi (t) = (a,b) + t(h_1,h_2) = (a+th_1, b+th_2) $ for $t \in \mathbb{R}$. Note $\phi(0)=(a,b)$ and $\phi '(t) = (h_1,h_2) = \phi ' (0)$. Construct $g = f \,{\scriptstyle \stackrel{\circ}{}}\, \phi : \mathbb{R} \rightarrow \mathbb{R}$ and choose $dom(g)$ such that $\phi(t) \in U$ for $t \in dom(g)$. This function $g$ is a real-valued function of a real variable and we will be able to apply Taylor's theorem from calculus II on $g$. However, to differentiate $g$ we'll need tools from calculus III to sort out the derivatives. In particular, as we differentiate $g$, note we use the chain rule for functions of several variables: $ \begin{align} \notag g'(t) =(f \,{\scriptstyle \stackrel{\circ}{}}\, \phi)'(t) &= f'(\phi(t))\phi '(t) \\ \notag &= \nabla f(\phi(t)) \cdot (h_1,h_2) \\ \notag &= h_1f_x(a+th_1, b+th_2)+h_2f_y(a+th_1, b+th_2) \notag \end{align} $ Note $g'(0)=h_1f_x(a, b)+h_2f_y(a, b)$. Differentiate again (I omit $(\phi(t))$ dependence in the last steps), $ \begin{align} \notag g''(t) &= h_1f_x '(a+th_1, b+th_2)+h_2f_y '(a+th_1, b+th_2) \\ \notag &= h_1 \nabla f_x (\phi(t)) \cdot (h_1,h_2) +h_2 \nabla f_y (\phi(t)) \cdot (h_1,h_2) \\ \notag &= h_1^2f_{xx}+ h_1h_2f_{yx} + h_2h_1f_{xy}+ h_2^2f_{yy} \\ \notag &= h_1^2f_{xx}+ 2h_1h_2f_{xy} + h_2^2f_{yy} \notag \end{align} $ Thus, making explicit the point dependence, $g''(0) = h_1^2f_{xx}(a,b)+ 2h_1h_2f_{xy}(a,b) + h_2^2f_{yy}(a,b)$. We may construct the Taylor series for $g$ up to quadratic terms: $ \begin{align} \notag g(0+t) &= g(0)+tg'(0)+\frac{1}{2}g''(0) + \cdots \\ \notag &= f(a,b)+t[h_1f_x(a, b)+h_2f_y(a, b)]+\frac{t^2}{2}\bigl[ h_1^2f_{xx}(a,b)+ 2h_1h_2f_{xy}(a,b) + h_2^2f_{yy}(a,b) \bigr] + \cdots \\ \notag \end{align} $ Note that $g(t)= f(a+th_1, b+th_2)$ hence $g(1) = f(a+h_1, b+h_2)$ and consequently, $ \begin{align} \notag f(a+h_1, b+h_2) &= f(a,b)+h_1f_x(a, b)+h_2f_y(a, b)+ \\ \notag & \qquad +\frac{1}{2}\biggl[ h_1^2f_{xx}(a,b)+ 2h_1h_2f_{xy}(a,b) + h_2^2f_{yy}(a,b) \biggr] + \cdots \notag \end{align} $ Omitting point dependence on the $2^{nd}$ derivatives, $ \boxed{ f(a+h_1, b+h_2) = f(a,b)+h_1f_x(a,b)+h_2f_y(a,b)+ \tfrac{1}{2}\bigl[ h_1^2f_{xx}+ 2h_1h_2f_{xy} + h_2^2f_{yy} \bigr] + \cdots }$ Sometimes we'd rather have an expansion about $(x,y)$. To obtain that formula simply substitute $x-a = h_1$ and $y-b = h_2$. Note that the point $(a,b)$ is fixed in this discussion so the derivatives are not modified in this substitution, $ \begin{align} \notag f(x, y) &= f(a,b)+(x-a)f_x(a, b)+(y-b)f_y(a, b)+ \\ \notag & \qquad +\frac{1}{2}\biggl[ (x-a)^2f_{xx}(a,b)+ 2(x-a)(y-b)f_{xy}(a,b) + (y-b)^2f_{yy}(a,b) \biggr] + \cdots \notag \end{align} $ At this point we ought to recognize the first three terms give the tangent plane to $z = f(z,y)$ at $(a,b,f(a,b))$. The higher order terms are nonlinear corrections to the linearization, these quadratic terms form a quadratic form. If we computed third, fourth or higher order terms we will find that, using $a=a_1$ and $b=a_2$ as well as $x=x_1$ and $y=x_2$, $ \boxed{f(x, y) = \sum_{n=0}^{\infty} \sum_{i_1=0}^{2}\sum_{i_2=0}^{2} \cdots \sum_{i_n=0}^{2} \frac{1}{n!} \frac{\partial^{(n)}f(a_1,a_2)}{\partial x_{i_1}\partial x_{i_2} \cdots \partial x_{i_n}} (x_{i_1} -a_{i_1})(x_{i_2} -a_{i_2})\cdots (x_{i_n} -a_{i_n})} $