Python3, 361 bytes

Optimized to work on larger obstacle courses. See TIO link for more test cases.

def f(b):
 q,s=[(0,0,b[0][0],[(0,0)])],[sum(b[0])+sum(i[-1]for i in b[1:]),sum(b[-1])+sum(i[0]for i in b[:-1])]
 for x,y,t,p in q:
  if(x,y)==(J:=len(b)-1,K:=len(b[0])-1):s+=[t];continue
  for X,Y in[(1,0),(-1,0),(0,1),(0,-1)]:
   n,m=x+X,y+Y
   if 0<=n<=J and 0<=m<=K and(n,m)not in p and(s==[]or t+b[n][m]<min(s)):q+=[(n,m,t+b[n][m],p+[(n,m)])]
 return min(s)

Try it online!

Matlab, 207 199 bytes

Given the matrix with all the times A, the idea is using a new B matrix that has every entry set to Inf except the initial entry. Then we generate a new matrix C where we apply following formula to each pair of coordinates (i,j)

 C(i,j) = min( B(i,j), B(i-1,j)+A(i,j), B(i+1,j)+A(i,j), B(i,j-1)+A(i,j), B(i,j+1)+A(i,j) );

This means: either B(i,j) already represents the minimum time to get to (i,j) or we can get to B(i,j) quicker via one of the neighbours.

Then we replace B with C and repeat this step enough times (until nothing changes anymore), and #rows + #columns is an upper bound for this.

After that, we just need to print the top right element (or in matrix-speak: bottom right) element.

This idea is basically stolen from the Floyd-Warshall algorithm.

Note that for doing this, I need also to consider some technical details, namely that I should not access elements outside the matrix, that is why I need the zero padding and shift of coordinates.

a=input('');
[r,c]=size(a);
A=padarray(a,[1,1]);
C=A+Inf;C(2,2)=A(2,2);
for i=1:r+c
    B=C;
    for i=2:r+1;
        for j=2:c+1;
            v=[-1,1];
            C(i,j)=min([B(i,j)-A(i,j),B(i+v,j)',B(i,j+v)]+A(i,j));
        end;
    end;
end;
disp(A(r+1,c+1))

Input: We need to reverse the order of the rows due to indexing, so the example given would be

[[2,6,1];[0,3,5];[0,1,2]]

or alternatively

[2,6,1;0,3,5;0,1,2]

So for this example we get following intermediate steps that will show the progression quite nicely (I removed the padding):

 2     Inf   Inf
 Inf   Inf   Inf
 Inf   Inf   Inf

 2     8     Inf
 2     Inf   Inf
 Inf   Inf   Inf

 2     8     9
 2     5     Inf
 2     Inf   Inf

 2     8     9
 2     5    10
 2     3    Inf

 2     8     9
 2     5    10
 2     3     5

PS: I love challenges like this one=)